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Chapter 5: Trigonometric Ratios Exercise – 5.1 Question: 1 Find the value of Trigonometric ratios in each of the following provided one of the six trigonometric ratios are given. (i) sin A = 2/3 (ii) cos A = 4/5 (iii) tan θ = 11/1 (iv) sin θ = 11/15 (v) tan α = 5/12 (vi) sin θ = √3/2 (vii) cos θ = 7/25 (viii) tan θ = 8/15 (ix) cot θ = 12/5 (x) sec θ = 13/5 (xi) cosec θ = √10 (xii) cos θ =12/15 Solution: (i) sin A = 2/3 Given: sin A = 2/3 … (1) By definition, By Comparing (1) and (2) We get, Perpendicular side = 2 and Hypotenuse = 3 Therefore, by Pythagoras theorem, AC^{2 }= AB^{2 +} BC^{2} Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB) Therefore, 3^{2} = AB^{2} + 2^{2} AB^{2 }= 3^{2} – 2^{2} AB^{2} = 9 – 4 AB^{2 }= 5 AB = √5 Hence, Base = √5 Therefore, Therefore, Therefore, cot A = √5/2 (ii) cos A = 4/5 Given: cos A = 4/5 … (1) By Definition, By comparing (1) and (2) We get, Base = 4 and Hypotenuse = 5 Therefore, By Pythagoras theorem, AC^{2 }= AB^{2 }+ BC^{2} Substituting the value of base (AB) and hypotenuse (AC) and get the perpendicular side (BC) 5^{2} = 4^{2}+ BC^{2} BC^{2} = 5^{2} – 4^{2} BC^{2 }= 25 – 16 BC^{2} = 9 BC = 3 Hence, Perpendicular side = 3 Now, Therefore, sin A = 3/5 Now, cosec A = 1/(sin A) Therefore, cosec A = 1/(sin A) Therefore, Therefore, Therefore, tan A = 3/4 Now, cot A = 1/(tan A) Therefore, By definition, By Comparing (1) and (2) We get, Base = 1 and Perpendicular side = 5 Therefore, By Pythagoras theorem, AC^{2} = AB^{2} + BC^{2} Substituting the value of base side (AB) and perpendicular side (BC) and get hypotenuse (AC) AC^{2} = 1^{2} + 11^{2} AC^{2} = 1 + 121 AC^{2 }= 122 AC = √122 Therefore, Sin θ = 11√122 Therefore, Therefore, Therefore, By definition, By Comparing (1) and (2) We get, Perpendicular Side = 11 and Hypotenuse = 15 Therefore, By Pythagoras theorem, AC^{2} = AB^{2} + BC^{2} Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB) 15^{2} = AB^{2} +11^{2} AB^{2} = 15^{2 }– 11^{2} AB^{2}= 225 – 121 AB^{2 }= 104 Therefore, Therefore, Therefore, Therefore, Therefore, (v) tan α = 5/12 Given: tan α = 5/12 … (1) By definition, By comparing (1) and (2) We get, Base = 12 and Perpendicular side = 5 Therefore, By Pythagoras theorem, AC^{2} = AB^{2} + BC^{2} Substituting the value of base side (AB) and the perpendicular side (BC) and gte hypotenuse (AC) AC^{2} = 12^{2} + 5^{2} AC^{2} = 144 + 25 AC^{2 }= 169 AC = 13 Hence Hypotenuse = 13 Therefore, sin α = 5/13 Therefore, Therefore, By definition, By comparing (1) and (2) We get, Perpendicular side = √3 Hypotenuse = 2 Therefore, By Pythagoras theorem, AC^{2} = AB^{2} + BC^{2} Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB) AB^{2} = 4 - 3 AB^{2} = 1 AB = 1 Hence Base = 1 Therefore, cos θ = 1/2 Now, cosec θ = 1/sin θ Therefore, Therefore, Therefore, Therefore, (vii) cos θ = 7/25 Given: cos θ = 7/25 … (1) By definition, By comparing (1) and (2) We get, Base = 7 and Hypotenuse = 25 Therefore By Pythagoras theorem, AC ^{2 }= AB^{2} + BC^{2} Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC) 25^{2} = 7^{2 }+ BC^{2} BC^{2} = 25^{2} – 7^{2} BC^{2} = 625 – 49 BC = 576 BC = √576 BC = 24 Hence, Perpendicular side = 24 Therefore, Sin θ = 24/25 Now, cosec θ = 1/sin θ Therefore, Therefore, Therefore, Therefore, By definition, By comparing (1) and (2) We get, Base = 15 and Perpendicular side = 8 Therefore, By Pythagoras theorem, AC^{2}= 15^{2} + 8^{2} AC^{2}= 225 + 64 AC^{2} = 289 AC = √289 AC = 17 Hence, Hypotenuse = 17 Therefore, sin θ = 8/17 Now, cosec θ = 1/sin θ Therefore, Therefore, cos θ =15/17 Now, sec θ = 1/cos θ Therefore, Therefore, (ix) cot θ = 12/5 Given: cot θ = 12/5 … (1) By definition, By comparing (1) and (2) We get, Base = 12 and Perpendicular side = 5 Therefore, By Pythagoras theorem, AC^{2 }= AB^{2} + BC^{2} Substituting the value of base side (AB) and perpendicular side (BC) and get the hypotenuse (AC) AC^{2} = 12^{2} + 5^{2} AC^{2 }= 144 + 25 AC^{2} = 169 AC = √169 AC = 13 Hence, Hypotenuse = 13 Therefore, sin θ = 5/13 Now, cosec θ = 1/sin θ Therefore, Therefore, cos θ = 12/13 Now, sec θ = 1/cos θ Therefore, Therefore, (x) sec θ = 13/5 Given: sec θ = 13/5… (1) By definition, By comparing (1) and (2) We get, Base = 5 Hypotenuse = 13 Therefore, By Pythagoras theorem, Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC) 13^{2} = 5^{2} + BC^{2} BC^{2} = 13^{2 }– 5^{2} BC^{2 }= 169 – 25 BC^{2 }= 144 BC = √144 BC = 12 Hence, Perpendicular side = 12 Therefore, sin θ = 12/13 Now, cosec θ = 1/sin θ Therefore, Therefore, Therefore, tan θ = 12/5 Now, cot θ = 1/tan θ Therefore, (xi) cosec θ = √10 Given: cosec θ = √10/1 … (1) By definition By comparing (1) and (2) We get, Perpendicular side = 1 and Hypotenuse = √10 Therefore, By Pythagoras theorem, AC^{2 }= AB^{2} + BC^{2} Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB) AB^{2 }= 10 - 1 AB = √9 AB = 3 Hence, Base side = 3 Therefore, Therefore, Therefore, cot θ = 3 (xii) cos θ = 12/15 Given: cos θ = 12/15 … (1) By definition, By comparing (1) and (2) We get, Base = 12 and Hypotenuse = 15 Therefore, By Pythagoras theorem, AC^{2} = AB^{2 }+ BC^{2} Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC) 15^{2} = 12^{2} + BC^{2} BC^{2} = 15^{2} – 12^{2} BC^{2} = 225 – 144 BC ^{2}= 81 BC = √81 BC = 9 Hence, Perpendicular side = 9 Therefore, Therefore, Therefore, Therefore, Therefore, Question: 2 In a ΔABC, right angled at B, AB – 24 cm, BC = 7cm, Determine (i) sin A, cos A (ii) sin C, cos C Solution: (i) The given triangle is below: Given: In ΔABC, AB = 24 cm BC = 7cm ∠ABC = 90^{°} To find: sin A, cos A In this problem, Hypotenuse side is unknown Hence we first find hypotenuse side by Pythagoras theorem By Pythagoras theorem, We get, AC^{2} = AB^{2} + BC^{2} AC^{2} = 24^{2} + 7^{2} AC^{2} = 576 + 49 AC^{2 }= 625 AC = √625 AC = 25 Hypotenuse = 25 By definition, By definition, Answer: (ii) The given triangle is below: Given: In Δ ABC, AB = 24 cm BC = 7cm ∠ABC = 90^{°} To find: sin C, cos C In this problem, Hypotenuse side is unknown Hence we first find hypotenuse side by Pythagoras theorem By Pythagoras theorem, We get, AC^{2} = AB^{2} + BC^{2} AC^{2} = 24^{2} + 7^{2} AC^{2} = 576 + 49 AC^{2 }= 625 AC = √625 AC = 25 Hypotenuse = 25 By definition, By definition, By definition, Question: 3 In the below figure, find tan P and cot R. Is tan P = cot R? To find, tan P, cot R Solution: In the given right angled ΔPQR, length of side OR is unknown Therefore, by applying Pythagoras theorem in ΔPQR We get, PR^{2} = PQ^{2} + QR^{2} Substituting the length of given side PR and PQ in the above equation 13^{2 }= 12^{2} + QR^{2} QR^{2} = 13^{2} – 12^{2} QR^{2} = 169 – 144 QR^{2}= 25 QR =√25 By definition, we know that, Also, by definition, we know that Comparing equation (1) ad (2), we come to know that that R.H.S of both the equation are equal. Therefore, L.H.S of both equations is also equal tan P = cot R Answer: Question: 4 If sin A = 9/41, Compute cos A and tan A. Solution: Given: sin A = 9/41 … (1) To find: cos A, tan A By definition, By comparing (1) and (2) We get, Perpendicular side = 9 and Hypotenuse = 41 Now using the perpendicular side and hypotenuse we can construct ΔABC as shown figure. Length of side AB is unknown is right angled ΔABC, To find the length of side AB, we use Pythagoras theorem, Therefore, by applying Pythagoras theorem in ΔABC, We get, AC^{2} = AB^{2} + BC^{2} 41^{2} = AB^{2} + 9^{2} AB^{2} = 41^{2} – 9^{2} AB^{2} = 168 – 81 AB= 1600 AB = √1600 AB = 40 Hence, length of side AB = 40 Now By definition, Now, By definition, Question: 5 Given 15 cot A = 8, find sin A and sec A. Solution: Given: 15 cot A = 8 To find: sin A, sec A Since 15 cot A =8 By taking 15 on R.H.S We get, Since 15 cot A = 8 By taking 15 on R.H.S We get, cot A = 8/15 By definition, cot A = 1/(tan A) Hence, Comparing equation (1) and (2) We get, Base side adjacent to ∠A = 8 Perpendicular side opposite to ∠A = 15 ΔABC can be drawn below using above information Hypotenuse side is unknown. Therefore, we find side AC of ΔABC by Pythagoras theorem. So, by applying Pythagoras theorem to ΔABC We get, AC^{2} = AB^{2} +BC^{2} Substituting values of sides from the above figure AC^{2} = 8^{2} + 15^{2} AC^{2} = 64 + 225 AC^{2} = 289 AC = √289 AC = 17 Therefore, hypotenuse =17 Now by definition, Substituting values of sides from the above figure Sin A = 15/17 By definition, sec A = 1/cos A Hence, Substituting values of sides from the above figure Sec A = 17/8 Answer: sin A =15/17, sec A = 17/8 Question: 6 In ΔPQR, right angled at Q, PQ = 4cm and RQ = 3 cm .Find the value of sin P, sin R, sec P and sec R. Solution: Given: ΔPQR is right angled at vertex Q. PQ = 4cm RQ = 3cm To find, sin P, sin R, sec P, sec R Given ΔPQR is as shown figure. Hypotenuse side PR is unknown. Therefore, we find side PR of ΔPQR by Pythagoras theorem By applying Pythagoras theorem to ΔPQR We get, PR^{2} = PQ^{2} + RQ^{2} Substituting values of sides from the above figure PR^{2} = 4^{2} +3^{2} PR^{2} = 16 + 9 PR^{2} = 25 PR = √25 PR = 5 Hence, Hypotenuse =5 Now by definition, sin P = RQ/PR Substituting values of sides from the above figure sin P = 3/5 Now by definition, sin R = PQ/PR Substituting the values of sides from above figure sin R = 4/5 By definition, sec P = 1/cos P Substituting values of sides from the above figure sec P = PR/PQ sec P = 5/4 By definition, sec R = 1/cos R Substituting values of sides from the above figure sec R = PR/RQ sec R = 5/3 Answer: sin P = 3/5, sin R = 4/5, sec P = 5/4, sec R = 5/3 Question: 7 If cot θ = 7/8, evaluate (ii) cot^{2 }θ cot^{2}θ Solution: Given: cot θ = 7/8 We know the following formula (a + b)(a – b) = a^{2} - b^{2} By applying the above formula in the numerator of equation (1) We get, (1 + sin θ) × (1 – sin θ) = 1 – sin^{2}θ .... (2) (Where, a = 1 and b = sin θ) Similarly, By applying formula (a + b) (a – b) = a^{2} – b^{2} in the denominator of equation (1). We get, (1 + cos θ)(1 – cos θ) = 1 – cos^{2 }θ (1 + cos θ)(1 – cos θ) = 1 – cos^{2}θ … (Where a = 1 and b = cos θ cos θ Substituting the value of numerator and denominator of equation (1) from equation (2), equation (3). Therefore, Since, cos^{2}θ + sin^{2} θ = 1 cos^{2} θ + sin^{2} θ = 1 Therefore, cos^{2} θ = 1 – sin^{2}Θcos^{2} θ =1 – sin^{2} θ Also, sin^{2} θ = 1 – cos^{2} θ sin^{2} θ = 1 – cos^{2} θ Putting the value of 1 – sin^{2} θ and 1 – cos^{2} θ in equation (4) We get, We know that, Since, it is given that cot θ = 7/8 Therefore, (ii) Given: cot θ = 7/8 To evaluate: cot^{2} θ cot θ =7/8 Squaring on both sides, We get, Answer: 49/64 Question: 8 If 3 cot A = 43 cot A = 4, check whether Solution: Given: 3 cot A = 4 To check whether cos^{2}A –sin^{2}A or not. 3 cot A = 4 Dividing by 3 on both sides, We get, cot A = 4/3 … (1) By definition, cot A = 1/tan A Therefore, Comparing (1) and (2) We get, Base side adjacent to ∠A = 4 Perpendicular side opposite to ∠A = 3 Hence ΔABC is as shown in figure. In ΔABC , Hypotenuse is unknown Hence, it can be found by using Pythagoras theorem Therefore by applying Pythagoras theorem in ΔABC We get AC^{2 }= AB^{2} + BC^{2} Substituting the values of sides from the above figure AC^{2} = 4^{2} + 3^{2} AC^{2} = 16 + 9 AC^{2} = 25 AC = √25 AC = 5 Hence, hypotenuse = 5 To check whethercos^{2}A – sin^{2}A or not. We get thee values of tan A, cos A, sin A By definition, tan A =1/(cot A) Substituting the value of cot A from equation (1) We get, tan A = 1/4 tan A = 3/4 …. (3) Now by definition, Substituting the values of sides from the above figure Now we first take L.H.S of equation Substituting value of tan A from equation (3) We get, Taking L.C.M on both numerator and denominator We get, Now we take R.H.S of equation whether R.H.S = cos^{2}A – sin^{2}A Substituting value of sin A and cos A from equation (4) and (5) We get, Comparing (6) and (7) We get. Answer: Question: 9 If tan θ = a/b, find the value of Solution: Given: tan θ = a/b … (1) Now, we know that Therefore equation (1) become as follows Now, by applying invertendo We get, Now by applying Componendo – dividendo We get, Therefore, Question: 10 If 3 tan θ = 4, find the value of Solution: Given: If 3 tan θ = 4/3 tan θ = 4 Therefore, tan θ = 4/3 … (1) Now, we know that Therefore equation (1) becomes Now, by applying Invertendo to equation (2) We get, Now, multiplying by 4 on both sides We get Therefore Now, multiplying by 2 on both sides of equation (3) We get, Now by applying componendo in above equation We get, Therefore, Therefore, on L.H.S sin θ sin θ cancels and we get, Therefore, 4 cos θ – sin θ = 44 cos θ – sin θ = 4 Question: 11 If 3 cot θ = 2, find the value of Solution: Given: 3 cot θ = 2 Therefore, Now, we know that Therefore equation (1) becomes Now, by applying invertendo to equation (2) Now, multiplying by 4/3 on both sides, We get, Therefore, 3 cancels out on R.H.S and We get, Now by applying invertendo dividendo in above equation We get, Now, multiplying by 2/6 on both sides of equation (3) We get, Therefore, 2 cancels out on R.H.S and We get, Now by applying componendo in above equation We get, Now, by dividing equation (4) by (5) We get, Therefore, Therefore, on L.H.S (3 sin θ) cancels out and we get, Now, by taking 2 in the numerator of L.H.S on the R.H.S We get, Therefore, 2 cancels out on R.H.S and We get, Hence answer, Question: 12 If tan θ = a/b, prove that Solution: Given: Now, we know that Therefore equation (1) becomes Now, by multiplying by a/b on both sides of equation (2) We get, Therefore, Now by applying dividendo in above equation (3) We get, Now by applying componendo in equation (4) We get, Now, by dividing equation (4) by equation (5) We get, Therefore, Therefore, b cos θ and b^{2 }cancels on L.H.S and R.H.S respectively Hence, it is proved that Question: 13 If sec θ = 13/5, show that Solution: Given: sec θ = 13/5 To show that Now, we know that Therefore, Therefore, Now, we know that Now, by comparing equation (1) and (2) We get, Base side adjacent to ∠θ = 5 And Hypotenuse =13 Therefore from above figure Base side BC = 5 Hypotenuse AC = 13 Side AB is unknown. It can be determined by using Pythagoras theorem Therefore by applying Pythagoras theorem We get, AC^{2} = AB^{2} + BC^{2} Therefore by substituting the values of known sides We get, 13^{2} = AB^{2} + 5^{2} Therefore, AB^{2}= 13^{2} – 5^{2} AB^{2}= 169 – 25 AB^{2} = 144 AB = √144 Therefore, AB = 12 …. (3) Now, we know that sin θ = AB/AC sin θ = 12/13 … (4) Now L.H.S of the equation to be proved is as follows Substituting the value cos θ of sin θ and from equation (1) and (4) respectively We get, Therefore, L.H.S = 3 Hence proved that, Question: 14 If cos θ = 12/13, show that sin θ(1 – tan θ) = 35/156 Solution: To show that Now we know that Therefore, by comparing equation (1) and (2) We get, Base side adjacent to ∠θ = 12 And Hypotenuse = 13 Therefore from above figure Base side BC = 12 Hypotenuse AC = 13 Side AB is unknown and it can be determined by using Pythagoras theorem Therefore by applying Pythagoras theorem We get, AC^{2}= AB^{2} + BC^{2} Therefore by substituting the values of known sides We get, 13^{2}= AB^{2} + 12^{2} Therefore, AB^{2}= 13^{2 }– 12^{2} AB^{2}= 169 – 144 AB = 25 AB = √25 AB = 5 …. (3) Now, we know that Now from figure (a) We get, sin θ = AB/AC Therefore, sin θ = 5/12… (5) Now L.H.S of the equation to be proved is as follows L.H.S of the equation to be proved is as follows L.H.S = sin θ (1 – tan θ] …. (6) Substituting the value of sin θ and tan θ from equation (4) and (5) We get, Taking L.C.M inside the bracket We get, Therefore, Now by opening the bracket and simplifying We get, From equation (6) and (7),it can be shown that Question: 15 Solution: Now, we know that Therefore, Therefore, Comparing Equation (1) and (2) We get. Base side adjacent to ∠θ = 1 Perpendicular side opposite to ∠θ = √3 Therefore, triangle representing angle √3 is as shown below Therefore, by substituting the values of known sides We get, Therefore, AC^{2} = 3 + 1 AC^{2 }= 4 AC = √4 Therefore, AC = 2 … (3) Now, we know that Now from figure (a) Therefore from figure (a) and equation (3), Now we know that Now from figure (a) We get, BC/AC Therefore from figure (a) and equation (3), Now, L.H.S of the equation to be proved is as follows Substituting the value of from equation (4) and (5) We get, Now by taking L.C.M in numerator as well as denominator We get, Therefore, Therefore, Question: 16 Solution: To show that Now, we know that Therefore, Comparing equation (1) and (2) We get. Perpendicular side opposite to ∠θ = 1 Base side adjacent to ∠θ = √7 Therefore, Triangle representing ∠ θ is shown figure. Hypotenuse AC is unknown and it can be found by using Pythagoras theorem Therefore by applying Pythagoras theorem We get, AC^{2 }= AB^{2} + BC^{2} Therefore by substituting the values of known sides We get, Therefore, AC ^{2} = 1 +7 AC^{2} = 8 AC = √8 Therefore, Now we know that Now, we know that Therefore, from equation (4) We get, Now, we know that Now from figure (a) We get, cos θ = BC/AC Therefore from figure (a) and equation (3) Now we know that Therefore, from equation (6) We get, Now, L.H.S of the equation to be proved is as follows Substituting the value of cosec θ and sec θ from equation (6) and (7) We get, Therefore, Therefore, L.H.S = 48/64 L.H.S = 3/4 = R.H.S Therefore, Hence proved that Question: 17 If sec θ = 5/4, find the value of Solution: To find the value of Now we know that Therefore, Therefore from equation (1) Also, we know that cos^{2}θ + sin^{2}θ = 1 Therefore, Substituting the value of cos θ cos θ from equation (2) We get, Therefore, Also, we know that sec^{2}θ =1 + tan^{2 }θ sec^{2 }θ = 1 + tan^{2 }θ Therefore, Therefore, tan θ = 3/4 … (4) Also, cot θ = 1/tan θ Therefore from equation (4) We get, cot θ = 4/3 … (5) Substituting the value of cos θ, cot θ and tan θ from the equation (2), (3), (4) and (5) respectively in the expression below We get, = 12/7 Therefore, Question: 18 If sin θ = 12/13, find the value of Solution: To find the value of Now, we know the following trigonometric identity cosec^{2 }θ = 1 + tan^{2}θ Therefore, by substituting the value of tan θ from equation (1) We get, By taking L.C.M on the R.H.S We get, Therefore Therefore Now, we know that Therefore Now, we know the following trigonometric identity cos^{2}θ + sin^{2}θ = 1 Therefore, cos^{2}θ = 1 – sin^{2}θ Now by substituting the value of sin θ from equation (3) We get, Therefore, by taking L.C.M on R.H.S We get, Now, by taking square root on both sides We get, Therefore, Substituting the value of sin θ and cos θ from equation (3) and (4) respectively in the equation below Therefore, Therefore, Question: 19 If cos θ = 3/5, find the value of Solution: To find the value of Now we know the following trigonometric identity cos^{2}θ + sin^{2}θ = 1 Therefore by substituting the value of cos θ from equation (1) We get, Therefore, Therefore by taking square root on both sides We get, Now, we know that Therefore by substituting the value of sin θ and cos θ from equation (2) and (1) respectively We get, Now, by substituting the value of sin θ and of tan θ from equation (2) and equation (4) respectively in the expression below We get, Therefore, Question: 20 If sin θ = 3/5, find the value of Solution: Given: Now, we know the following trigonometric identity cos^{2}θ + sin^{2}θ = 1 Therefore by substituting the value of cos θ from equation (1) We get, Therefore, Now by taking L.C.M We get, Therefore, by taking square roots on both sides We get, cos θ = 4/5 Therefore, cos θ = 4/5 … (2) Now we know that tan θ = sin θ/cos θ Therefore by substituting the value of sin θ and cos θ from equation (1) and (2) respectively We get, Also, we know that Therefore from equation (3) We get, Now by substituting the value of cos θ, tan θ and cot θ from equation (2), (3) and (4) respectively from the expression below Therefore, Question: 21 If tan θ = 24/7, find that sin θ + cos θ Solution: Given: tan θ =24/7 … (1) To find, Sin θ + cos θ Now we know that tan θ is defined as follows Now by comparing equation (1) and (2) We get, Perpendicular side opposite to ∠θ = 24 Base side adjacent to ∠θ = 7 Therefore triangle representing ∠θ is as shown figure. Side AC is unknown and can be found by using Pythagoras theorem Therefore, AC^{2 }= AB^{2} + BC^{2} Now by substituting the value of unknown sides from figure We get, AC^{2}= 24^{2} +7^{2} AC = 576 + 49 AC = 625 Now by taking square root on both sides, We get, AC = 25 Therefore, Hypotenuse side AC = 25 …. (3) Now we know sin θ is defined as follows Therefore from figure (a) and equation (3) We get, Now we know that cos θ is defined as follows Therefore by substituting the value of sin θ and cos θ from equation (4) and (5) respectively, we get Question: 21 If sin θ = a/b, find sec θ + tan θ in terms of a and b. Solution: Given: sin θ = a/b … (1) To find: sec θ + tan θ Now we know, sin θ is defined as follows Now by comparing equation (1) and (2) We get, Perpendicular side opposite to ∠θ = a Hypotenuse = b Therefore triangle representing ∠θ is as shown figure. Hence side BC is unknown Now we find BC by applying Pythagoras theorem to right angled ΔABC Therefore, AC^{2} = AB^{2} + BC^{2} Now by substituting the value of sides AB and AC from figure (a) We get, b^{2} = a^{2} + BC^{2} Therefore, BC^{2} = b^{2} – a^{2} Now by taking square root on both sides We get, Therefore, Now we know cos θ is defined as follows Therefore from figure (a) and equation (3) We get, Now we know, Therefore, Now we know, Now by substituting the values from equation (1) and (3) We get, Therefore, Now we need to find sec θ + tan θ Now by substituting the values of sec θ and tan θ from equation (5) and (6) respectively We get, We get, Now by substituting the value in above expression We get, Now,present in the numerator as well as denominator of above denominator of above expression gets cancels we get, Square root is present in the numerator as well as denominator of above expression. Therefore we can place both numerator and denominator under a common square root sign Therefore, Question: 23 If 8 tan A = 15, find sin A – cos A Solution: Given: 8 tan A = 15 Therefore, To find: sin A – cos A Now we know tan A is defined as follows Now by comparing equation (1) and (2) We get Perpendicular side opposite to ∠A = 15 Base side adjacent to ∠A = 8 Therefore triangle representing angle A is as shown figure. Side AC = is unknown and can be found by using Pythagoras theorem Therefore, AC^{2 }= AB^{2} + BC^{2} Now by substituting the value of known sides from figure (a) We get, AC^{2 }= 15^{2} + 8^{2} AC^{2} = 225 + 64 AC = 289 Now by taking square root on both sides We get, AC = √289 AC = 17 Therefore Hypotenuse side AC = 17 … (3) Now we know, sin A is defined as follows Therefore from figure (a) and equation (3) We get, Now we know, cos A is defined as follows Therefore from figure (a) and equation (3) We get, Now we find the value of expression sin A – cos A Therefore by substituting the value the value of sin A and cos A from equation (4) and (5) respectively, we get, Question: 24 If tan θ = 20/21, show that Solution: Given: tan θ = 20/21 Now we know that Therefore, tan θ = 20/21 Side AC be the hypotenuse and can be found by applying Pythagoras theorem Therefore, AC^{2 }= AB^{2} + BC^{2} AC^{2}= 21^{2} + 20^{2} AC^{2} = 441 + 400 AC^{2} = 841 Now by taking square root on both sides We get, AC = √841 AC = 29 Therefore Hypotenuse side AC = 29 Now we know, sin θ is defined as follows, Therefore from figure and above equation We get, Now we know cos θ is defined as follows Therefore from figure and above equation We get, Now we need to find the value of expression Therefore by substituting the value of sin θ and cos θ from above equations, we get Therefore after evaluating we get, Hence, Question: 25 If cosec A = 2, find Solution: Given: cosec A = 2 Here BC is the adjacent side, By applying Pythagoras theorem, AC^{2 }= AB^{2} + BC^{2} 4 = 1 + BC^{2} BC^{2 }= 3 BC = √3 Now we know that Substitute all the values of sin A, cos A and tan A from the equations (1), (2) and (3) respectively We get. = 2 Hence, Question: 26 If ∠A and ∠B are acute angles such that cos A =cos B, then show that ∠A = ∠B Solution: Given: ∠A and ∠B are acute angles cos A = cos B such that ∠A = ∠B Let us consider right angled triangle ACB Now since cos A = cos B Therefore Now observe that denominator of above equality is same that is AB Therefore AC = BC We know that when two sides of triangle are equal, then opposite of the sides are also Equal. Therefore We can say that Angle opposite to side AC = angle opposite to side BC Therefore, ∠B = ∠A Hence, ∠A = ∠B Question: 27 In a ΔABC, right angled triangle at A, if tan C = √3, find the value of sin B cos C + cos B sin C. Solution: Given: ΔABC To find: sin B cos C + cos B sin C The given a ΔABC is as shown in figure Side BC is unknown and can be found using Pythagoras theorem, Therefore, BC^{2 }= AB^{2} + AC^{2} BC^{2 }= 3 +1 BC^{2} = 4 Now by taking square root on both sides We get, BC = √4 BC = 2 Therefore Hypotenuse side BC = 2 … (1) Now, Therefore, Now by substituting the values from equation (1) and figure We get, Now, Therefore, Now substituting the value from equation Similarly Now by definition, So by evaluating Now, by substituting the value of sin B, cos B, sin C and cos C from equation (2), (3), (4) and (5) respectively in sin B cos C + cos B sin C Sin B cos C + cos B sin C = 1 Hence, sin B cos C + cos B sin C = 1 Question: 28 State whether the following are true or false. Justify your answer. (i) The value of tan A is always less than 1. (ii) sec A = 12/5 for some value of ∠A. (iii) cos A is the abbreviation used for the cosecant of ∠A. (iv) sin θ = 4/3 for some angle θ. Solution: (i) tan A < 1 Value of tan A at 45^{°} i.e… tan 45 = 1 As value as A increases to 90^{°} Tan A becomes infinite So given statement is false. (ii) sec A = 12/5 for some value of angle if M-I sec A = 2.4 sec A > 1 So given statements is true. M- II For sec A = 12/5 we get adjacent side = 13 Subtending 9i at B. So, given statement is true. (iii) Cos A is the abbreviation used for cosecant of angle A. The given statement is false. As such cos A is the abbreviation used for cos of angle A, not as cosecant of angle A. (iv) cot A is the product of cot A and A Given statement is false? cot A is a co-tangent of angle A and co-tangent of angle (v) sin θ = 4/3 for some angle θ. Given statement is false Since value of sin θ is less than (or) equal to one. Here value of sin θ exceeds one, So given statement is false. Question: 29 Solution: As shown in figure Here BC is the adjacent side, By applying Pythagoras theorem, AC^{2 }= AB^{2 }+ BC^{2} 169 = 144 + BC^{2} BC^{2 }= 169 – 144 BC^{2 }= 25 BC = 5 Now we know that, We also know that, tan θ = sin θ/cos θ Therefore, substituting the value of sin θ and cos θ from above equations We get, tan θ = 12/5 Now substitute all the values of sin θ, cos θ and tan θ from above equations in We get, Therefore by further simplifying we get, Therefore, Hence, Question: 30 If cos θ = 5/13, find the value of Solution: Given: If cos θ = 5/13 ….. (1) To find: The value of expression Now we know that Now when we compare equation (1) and (2) We get, Base side adjacent to ∠θ = 5 Hypotenuse = 13 Therefore, Triangle representing ∠θ is as shown figure. Perpendicular side AB is unknown and it can be found by using Pythagoras theorem Therefore by applying Pythagoras theorem We get, AC^{2}= AB^{2} + BC^{2} Therefore by substituting the values of known sides, AB^{2} = 13^{2} – 5^{2} AB^{2}= 169 -25 AB^{2} = 144 AB = 12 …. (3) Now we know from figure and equation, Now we know that, Now w substitute all the values from equation (1), (4) and (5) in the expression below, Therefore We get, Therefore by further simplifying we get, Therefore, Hence, Question: 31 Solution: Given: sec A = 17/8 Now we know that Now, by substituting the value of sec A We get, Now we also know that, sin^{2}A + cos^{2}A = 1 Therefore sin^{2}A = 1 – cos^{2}A Now by taking square root on both sides, We get, sin A = 15/17 We also know that, Now by substituting the value of all the terms, We get, tan A = 15/8 Now from the expression of above equation which we want to prove: Now by substituting the value of cos A ad sin A from equation (3) and (4) We get, From expression Now by substituting the value of tan A from above equation We get, Therefore, We can see that, Question: 32 If sin θ = 3/4, Prove that Solution: To prove: By definition, By comparing (1) and (3) We get, Perpendicular side = 3 and Hypotenuse = 4 Side BC is unknown. So we find BC by applying Pythagoras theorem to right angled ΔABC Hence, AC^{2} = AB^{2} + BC^{2} Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC) Therefore, 4^{2 }= 3^{2} + BC^{2} BC^{2} = 16 – 9 BC^{2 }= 7 BC = √7 Hence, Base side BC = √7 … (3) Now cos A = BC/AC √7/4 … (4) Now, cosec A = 1/(sin A) Therefore, from fig and equation (1) cosec A = 4/3 … (5) Now, similarly Sec A = 4/√7 … (6) Further we also know that Therefore by substituting the values from equation (1) and (4), We get, Now by substituting the value of cosec A, sec A and cot A from the equations (5), (6), and (7) in the L.H.S of expression (2) Hence it is proved that, Question: 33 If sec A = 17/8, Verify that Solution: Given: sec A = 17/8 … (1) To verify: Now we know that sec A = 1/(cos A) Therefore, cos A = 1/(sec A) We get, cos A = 8/17 … (3) Similarly we can also get, sin A = 15/17 …. (4) An also we know that tan A = (sin A)/(cos A) tan A = 15/8 … (5) Now from the expression of equation (2) L.H.S: Missing close Now by substituting the value of cos A and sin A from equation (3) and (4) We get, Now by substituting the value of tan A from equation (5) We get, Now by comparing equation (6) and (7) We get, Question: 34 If cot θ = 3/4, prove that Solution: Given: cot θ = 3/4 Prove that: Now we know that Here AC is the hypotenuse and we can find that by applying Pythagoras theorem AC^{2 }= AB^{2} +BC^{2} AC^{2} = 16 +9 AC^{2 }= 25 AC = 5 Similarly Now on substituting the values in equations we get, Therefore, Question: 35 If 3 cos θ – 4 sin θ = 2 cos θ + sin θ, find tan θ tan θ Solution: Given: 3 cos θ - 4 sin θ = 2 cos θ + sin θ To find: tan θ tan θ We can write this as: 3 cos θ – 4 sin θ = 2 cos θ + sin θ cos θ = 5 sin θ Dividing both the sides by cos θ, We get, 1 = 5 tan θ tan θ = 1 Hence, tan θ = 1 Question: 36 If ∠A and ∠P are acute angles such that tan A = tan P, then show ∠A = ∠P Solution: Given: A and P are acute angles tan A = tan P Prove that: ∠A = ∠P Let us consider right angled triangle ACP We know ∴ tan A = tan P PC = AC [∵ Angle opposite to equal sides are equal] ∠A = ∠P
Find the value of Trigonometric ratios in each of the following provided one of the six trigonometric ratios are given.
(i) sin A = 2/3
(ii) cos A = 4/5
(iii) tan θ = 11/1
(iv) sin θ = 11/15
(v) tan α = 5/12
(vi) sin θ = √3/2
(vii) cos θ = 7/25
(viii) tan θ = 8/15
(ix) cot θ = 12/5
(x) sec θ = 13/5
(xi) cosec θ = √10
(xii) cos θ =12/15
Given: sin A = 2/3 … (1)
By definition,
By Comparing (1) and (2)
We get,
Perpendicular side = 2 and Hypotenuse = 3
Therefore, by Pythagoras theorem,
AC^{2 }= AB^{2 +} BC^{2}
Substituting the value of perpendicular side (BC) and hypotenuse (AC) and get the base side (AB)
Therefore,
3^{2} = AB^{2} + 2^{2}
AB^{2 }= 3^{2} – 2^{2}
AB^{2} = 9 – 4
AB^{2 }= 5
AB = √5
Hence, Base = √5
cot A = √5/2
Given: cos A = 4/5 … (1)
By Definition,
By comparing (1) and (2)
Base = 4 and
Hypotenuse = 5
By Pythagoras theorem,
AC^{2 }= AB^{2 }+ BC^{2}
Substituting the value of base (AB) and hypotenuse (AC) and get the perpendicular side (BC)
5^{2} = 4^{2}+ BC^{2}
BC^{2} = 5^{2} – 4^{2}
BC^{2 }= 25 – 16
BC^{2} = 9
BC = 3
Hence, Perpendicular side = 3
Now,
sin A = 3/5
Now, cosec A = 1/(sin A)
cosec A = 1/(sin A)
tan A = 3/4
Now, cot A = 1/(tan A)
Base = 1 and
Perpendicular side = 5
AC^{2} = AB^{2} + BC^{2}
Substituting the value of base side (AB) and perpendicular side (BC) and get hypotenuse (AC)
AC^{2} = 1^{2} + 11^{2}
AC^{2} = 1 + 121
AC^{2 }= 122
AC = √122
Sin θ = 11√122
Perpendicular Side = 11 and
Hypotenuse = 15
15^{2} = AB^{2} +11^{2}
AB^{2} = 15^{2 }– 11^{2}
AB^{2}= 225 – 121
AB^{2 }= 104
Given: tan α = 5/12 … (1)
Base = 12 and
Substituting the value of base side (AB) and the perpendicular side (BC) and gte hypotenuse (AC)
AC^{2} = 12^{2} + 5^{2}
AC^{2} = 144 + 25
AC^{2 }= 169
AC = 13
Hence Hypotenuse = 13
sin α = 5/13
Perpendicular side = √3
Hypotenuse = 2
AB^{2} = 4 - 3
AB^{2} = 1
AB = 1
Hence Base = 1
cos θ = 1/2
Now, cosec θ = 1/sin θ
Given: cos θ = 7/25 … (1)
Base = 7 and
Hypotenuse = 25
Therefore
AC ^{2 }= AB^{2} + BC^{2}
Substituting the value of base side (AB) and hypotenuse (AC) and get the perpendicular side (BC)
25^{2} = 7^{2 }+ BC^{2}
BC^{2} = 25^{2} – 7^{2}
BC^{2} = 625 – 49
BC = 576
BC = √576
BC = 24
Hence, Perpendicular side = 24
Sin θ = 24/25
Base = 15 and Perpendicular side = 8
AC^{2}= 15^{2} + 8^{2}
AC^{2}= 225 + 64
AC^{2} = 289
AC = √289
AC = 17
Hence, Hypotenuse = 17
sin θ = 8/17
cos θ =15/17
Now, sec θ = 1/cos θ
Given: cot θ = 12/5 … (1)
AC^{2 }= AB^{2} + BC^{2}
Substituting the value of base side (AB) and perpendicular side (BC) and get the hypotenuse (AC)
AC^{2 }= 144 + 25
AC^{2} = 169
AC = √169
Hence, Hypotenuse = 13
sin θ = 5/13
cos θ = 12/13
Given: sec θ = 13/5… (1)
Base = 5
Hypotenuse = 13
13^{2} = 5^{2} + BC^{2}
BC^{2} = 13^{2 }– 5^{2}
BC^{2 }= 169 – 25
BC^{2 }= 144
BC = √144
BC = 12
Hence, Perpendicular side = 12
sin θ = 12/13
tan θ = 12/5
Now, cot θ = 1/tan θ
Given: cosec θ = √10/1 … (1)
By definition
Perpendicular side = 1 and
Hypotenuse = √10
AB^{2 }= 10 - 1
AB = √9
AB = 3
Hence, Base side = 3
cot θ = 3
(xii) cos θ = 12/15
Given: cos θ = 12/15 … (1)
Base = 12 and Hypotenuse = 15
AC^{2} = AB^{2 }+ BC^{2}
15^{2} = 12^{2} + BC^{2}
BC^{2} = 15^{2} – 12^{2}
BC^{2} = 225 – 144
BC ^{2}= 81
BC = √81
BC = 9
Hence, Perpendicular side = 9
In a ΔABC, right angled at B, AB – 24 cm, BC = 7cm, Determine
(i) sin A, cos A
(ii) sin C, cos C
(i) The given triangle is below:
Given: In ΔABC, AB = 24 cm
BC = 7cm
∠ABC = 90^{°}
To find: sin A, cos A
In this problem, Hypotenuse side is unknown
Hence we first find hypotenuse side by Pythagoras theorem
AC^{2} = 24^{2} + 7^{2}
AC^{2} = 576 + 49
AC^{2 }= 625
AC = √625
AC = 25
Answer:
(ii) The given triangle is below:
Given: In Δ ABC, AB = 24 cm
To find: sin C, cos C
In the below figure, find tan P and cot R. Is tan P = cot R? To find, tan P, cot R
In the given right angled ΔPQR, length of side OR is unknown
Therefore, by applying Pythagoras theorem in ΔPQR
PR^{2} = PQ^{2} + QR^{2}
Substituting the length of given side PR and PQ in the above equation
13^{2 }= 12^{2} + QR^{2}
QR^{2} = 13^{2} – 12^{2}
QR^{2} = 169 – 144
QR^{2}= 25
QR =√25
By definition, we know that,
Also, by definition, we know that
Comparing equation (1) ad (2), we come to know that that R.H.S of both the equation are equal.
Therefore, L.H.S of both equations is also equal
tan P = cot R
If sin A = 9/41, Compute cos A and tan A.
Given: sin A = 9/41 … (1)
To find: cos A, tan A
Perpendicular side = 9 and Hypotenuse = 41
Now using the perpendicular side and hypotenuse we can construct ΔABC as shown figure.
Length of side AB is unknown is right angled ΔABC,
To find the length of side AB, we use Pythagoras theorem,
Therefore, by applying Pythagoras theorem in ΔABC,
41^{2} = AB^{2} + 9^{2}
AB^{2} = 41^{2} – 9^{2}
AB^{2} = 168 – 81
AB= 1600
AB = √1600
AB = 40
Hence, length of side AB = 40
Now
Given 15 cot A = 8, find sin A and sec A.
Given: 15 cot A = 8
To find: sin A, sec A
Since 15 cot A =8
By taking 15 on R.H.S
Since 15 cot A = 8
cot A = 8/15
cot A = 1/(tan A)
Hence,
Comparing equation (1) and (2)
Base side adjacent to ∠A = 8
Perpendicular side opposite to ∠A = 15
ΔABC can be drawn below using above information
Hypotenuse side is unknown.
Therefore, we find side AC of ΔABC by Pythagoras theorem.
So, by applying Pythagoras theorem to ΔABC
AC^{2} = AB^{2} +BC^{2}
Substituting values of sides from the above figure
AC^{2} = 8^{2} + 15^{2}
AC^{2} = 64 + 225
Therefore, hypotenuse =17
Now by definition,
Sin A = 15/17
sec A = 1/cos A
Sec A = 17/8
sin A =15/17, sec A = 17/8
In ΔPQR, right angled at Q, PQ = 4cm and RQ = 3 cm .Find the value of sin P, sin R, sec P and sec R.
Given: ΔPQR is right angled at vertex Q.
PQ = 4cm
RQ = 3cm
To find,
sin P, sin R, sec P, sec R
Given ΔPQR is as shown figure.
Hypotenuse side PR is unknown.
Therefore, we find side PR of ΔPQR by Pythagoras theorem
By applying Pythagoras theorem to ΔPQR
PR^{2} = PQ^{2} + RQ^{2}
PR^{2} = 4^{2} +3^{2}
PR^{2} = 16 + 9
PR^{2} = 25
PR = √25
PR = 5
Hence, Hypotenuse =5
sin P = RQ/PR
sin P = 3/5
sin R = PQ/PR
Substituting the values of sides from above figure
sin R = 4/5
sec P = 1/cos P
sec P = PR/PQ
sec P = 5/4
sec R = 1/cos R
sec R = PR/RQ
sec R = 5/3
sin P = 3/5,
sin R = 4/5,
sec P = 5/4,
If cot θ = 7/8, evaluate
(ii) cot^{2 }θ cot^{2}θ
Given: cot θ = 7/8
We know the following formula
(a + b)(a – b) = a^{2} - b^{2}
By applying the above formula in the numerator of equation (1)
(1 + sin θ) × (1 – sin θ) = 1 – sin^{2}θ .... (2) (Where, a = 1 and b = sin θ)
Similarly,
By applying formula (a + b) (a – b) = a^{2} – b^{2} in the denominator of equation (1).
(1 + cos θ)(1 – cos θ) = 1 – cos^{2 }θ (1 + cos θ)(1 – cos θ) = 1 – cos^{2}θ … (Where a = 1 and b = cos θ cos θ
Substituting the value of numerator and denominator of equation (1) from equation (2), equation (3).
Since,
cos^{2}θ + sin^{2} θ = 1 cos^{2} θ + sin^{2} θ = 1
cos^{2} θ = 1 – sin^{2}Θcos^{2} θ =1 – sin^{2} θ
Also, sin^{2} θ = 1 – cos^{2} θ sin^{2} θ = 1 – cos^{2} θ
Putting the value of 1 – sin^{2} θ and 1 – cos^{2} θ in equation (4)
We know that,
Since, it is given that cot θ = 7/8
(ii) Given: cot θ = 7/8
To evaluate: cot^{2} θ
cot θ =7/8
Squaring on both sides,
49/64
If 3 cot A = 43 cot A = 4, check whether
Given: 3 cot A = 4
To check whether cos^{2}A –sin^{2}A or not.
3 cot A = 4
Dividing by 3 on both sides,
cot A = 4/3 … (1)
cot A = 1/tan A
Comparing (1) and (2)
Base side adjacent to ∠A = 4
Perpendicular side opposite to ∠A = 3
Hence ΔABC is as shown in figure.
In ΔABC , Hypotenuse is unknown
Hence, it can be found by using Pythagoras theorem
Therefore by applying Pythagoras theorem in ΔABC
We get
Substituting the values of sides from the above figure
AC^{2} = 4^{2} + 3^{2}
AC^{2} = 16 + 9
AC^{2} = 25
AC = √25
AC = 5
Hence, hypotenuse = 5
To check whethercos^{2}A – sin^{2}A or not.
We get thee values of tan A, cos A, sin A
tan A =1/(cot A)
Substituting the value of cot A from equation (1)
tan A = 1/4
tan A = 3/4 …. (3)
Now we first take L.H.S of equation
Substituting value of tan A from equation (3)
Taking L.C.M on both numerator and denominator
Now we take R.H.S of equation whether
R.H.S = cos^{2}A – sin^{2}A
Substituting value of sin A and cos A from equation (4) and (5)
Comparing (6) and (7)
We get.
If tan θ = a/b, find the value of
Given: tan θ = a/b … (1)
Now, we know that
Therefore equation (1) become as follows
Now, by applying invertendo
Now by applying Componendo – dividendo
If 3 tan θ = 4, find the value of
Given: If 3 tan θ = 4/3 tan θ = 4
tan θ = 4/3 … (1)
Therefore equation (1) becomes
Now, by applying Invertendo to equation (2)
Now, multiplying by 4 on both sides
Now, multiplying by 2 on both sides of equation (3)
Now by applying componendo in above equation
Therefore, on L.H.S sin θ sin θ cancels and we get,
4 cos θ – sin θ = 44 cos θ – sin θ = 4
If 3 cot θ = 2, find the value of
Given: 3 cot θ = 2
Now, by applying invertendo to equation (2)
Now, multiplying by 4/3 on both sides,
Therefore, 3 cancels out on R.H.S and
Now by applying invertendo dividendo in above equation
Now, multiplying by 2/6 on both sides of equation (3)
Therefore, 2 cancels out on R.H.S and
Now, by dividing equation (4) by (5)
Therefore, on L.H.S (3 sin θ) cancels out and we get,
Now, by taking 2 in the numerator of L.H.S on the R.H.S
Hence answer,
If tan θ = a/b, prove that
Given:
Now, by multiplying by a/b on both sides of equation (2)
Now by applying dividendo in above equation (3)
Now by applying componendo in equation (4)
Now, by dividing equation (4) by equation (5)
Therefore, b cos θ and b^{2 }cancels on L.H.S and R.H.S respectively
Hence, it is proved that
If sec θ = 13/5, show that
Given: sec θ = 13/5
To show that
Now, by comparing equation (1) and (2)
Base side adjacent to ∠θ = 5
And
Hypotenuse =13
Therefore from above figure
Base side BC = 5
Hypotenuse AC = 13
Side AB is unknown. It can be determined by using Pythagoras theorem
Therefore by applying Pythagoras theorem
Therefore by substituting the values of known sides
13^{2} = AB^{2} + 5^{2}
AB^{2}= 13^{2} – 5^{2}
AB^{2}= 169 – 25
AB^{2} = 144
AB = √144
AB = 12 …. (3)
sin θ = AB/AC
sin θ = 12/13 … (4)
Now L.H.S of the equation to be proved is as follows
Substituting the value cos θ of sin θ and from equation (1) and (4) respectively
L.H.S = 3
Hence proved that,
If cos θ = 12/13, show that sin θ(1 – tan θ) = 35/156
Now we know that
Therefore, by comparing equation (1) and (2)
Base side adjacent to ∠θ = 12
Base side BC = 12
Side AB is unknown and it can be determined by using Pythagoras theorem
AC^{2}= AB^{2} + BC^{2}
13^{2}= AB^{2} + 12^{2}
AB^{2}= 13^{2 }– 12^{2}
AB^{2}= 169 – 144
AB = 25
AB = √25
AB = 5 …. (3)
Now from figure (a)
sin θ = 5/12… (5)
L.H.S of the equation to be proved is as follows
L.H.S = sin θ (1 – tan θ] …. (6)
Substituting the value of sin θ and tan θ from equation (4) and (5)
Taking L.C.M inside the bracket
Now by opening the bracket and simplifying
From equation (6) and (7),it can be shown that
Comparing Equation (1) and (2)
Base side adjacent to ∠θ = 1
Perpendicular side opposite to ∠θ = √3
Therefore, triangle representing angle √3 is as shown below
Therefore, by substituting the values of known sides
AC^{2} = 3 + 1
AC^{2 }= 4
AC = √4
AC = 2 … (3)
Therefore from figure (a) and equation (3),
We get, BC/AC
Now, L.H.S of the equation to be proved is as follows
Substituting the value of from equation (4) and (5)
Now by taking L.C.M in numerator as well as denominator
Perpendicular side opposite to ∠θ = 1
Base side adjacent to ∠θ = √7
Therefore, Triangle representing ∠ θ is shown figure.
Hypotenuse AC is unknown and it can be found by using Pythagoras theorem
AC ^{2} = 1 +7
AC^{2} = 8
AC = √8
Therefore, from equation (4)
cos θ = BC/AC
Therefore from figure (a) and equation (3)
Therefore, from equation (6)
Substituting the value of cosec θ and sec θ from equation (6) and (7)
L.H.S = 48/64
L.H.S = 3/4 = R.H.S
Hence proved that
If sec θ = 5/4, find the value of
To find the value of
Therefore from equation (1)
Also, we know that cos^{2}θ + sin^{2}θ = 1
Substituting the value of cos θ cos θ from equation (2)
Also, we know that
sec^{2}θ =1 + tan^{2 }θ sec^{2 }θ = 1 + tan^{2 }θ
tan θ = 3/4 … (4)
Also, cot θ = 1/tan θ
Therefore from equation (4)
cot θ = 4/3 … (5)
Substituting the value of cos θ, cot θ and tan θ from the equation (2), (3), (4) and (5) respectively in the expression below
= 12/7
If sin θ = 12/13, find the value of
Now, we know the following trigonometric identity
cosec^{2 }θ = 1 + tan^{2}θ
Therefore, by substituting the value of tan θ from equation (1)
By taking L.C.M on the R.H.S
cos^{2}θ + sin^{2}θ = 1
cos^{2}θ = 1 – sin^{2}θ
Now by substituting the value of sin θ from equation (3)
Therefore, by taking L.C.M on R.H.S
Now, by taking square root on both sides
Substituting the value of sin θ and cos θ from equation (3) and (4) respectively in the equation below
If cos θ = 3/5, find the value of
Now we know the following trigonometric identity
Therefore by substituting the value of cos θ from equation (1)
Therefore by taking square root on both sides
Therefore by substituting the value of sin θ and cos θ from equation (2) and (1) respectively
Now, by substituting the value of sin θ and of tan θ from equation (2) and equation (4) respectively in the expression below
If sin θ = 3/5, find the value of
Now by taking L.C.M
Therefore, by taking square roots on both sides
cos θ = 4/5
cos θ = 4/5 … (2)
tan θ = sin θ/cos θ
Therefore by substituting the value of sin θ and cos θ from equation (1) and (2) respectively
Therefore from equation (3)
Now by substituting the value of cos θ, tan θ and cot θ from equation (2), (3) and (4) respectively from the expression below
If tan θ = 24/7, find that sin θ + cos θ
Given: tan θ =24/7 … (1)
To find, Sin θ + cos θ
Now we know that tan θ is defined as follows
Now by comparing equation (1) and (2)
Perpendicular side opposite to ∠θ = 24
Base side adjacent to ∠θ = 7
Therefore triangle representing ∠θ is as shown figure.
Side AC is unknown and can be found by using Pythagoras theorem
Now by substituting the value of unknown sides from figure
AC^{2}= 24^{2} +7^{2}
AC = 576 + 49
AC = 625
Now by taking square root on both sides,
Hypotenuse side AC = 25 …. (3)
Now we know sin θ is defined as follows
Now we know that cos θ is defined as follows
Therefore by substituting the value of sin θ and cos θ from equation (4) and (5) respectively, we get
If sin θ = a/b, find sec θ + tan θ in terms of a and b.
sin θ = a/b … (1)
To find: sec θ + tan θ
Now we know, sin θ is defined as follows
Perpendicular side opposite to ∠θ = a
Hypotenuse = b
Hence side BC is unknown
Now we find BC by applying Pythagoras theorem to right angled ΔABC
Now by substituting the value of sides AB and AC from figure (a)
b^{2} = a^{2} + BC^{2}
BC^{2} = b^{2} – a^{2}
Now by taking square root on both sides
Now we know cos θ is defined as follows
Now we know,
Now by substituting the values from equation (1) and (3)
Now we need to find sec θ + tan θ
Now by substituting the values of sec θ and tan θ from equation (5) and (6) respectively
Now by substituting the value in above expression
Now,present in the numerator as well as denominator of above denominator of above expression gets cancels we get,
Square root is present in the numerator as well as denominator of above expression. Therefore we can place both numerator and denominator under a common square root sign
If 8 tan A = 15, find sin A – cos A
8 tan A = 15
To find: sin A – cos A
Now we know tan A is defined as follows
Therefore triangle representing angle A is as shown figure.
Side AC = is unknown and can be found by using Pythagoras theorem
Now by substituting the value of known sides from figure (a)
AC^{2 }= 15^{2} + 8^{2}
AC^{2} = 225 + 64
AC = 289
Therefore Hypotenuse side AC = 17 … (3)
Now we know, sin A is defined as follows
Now we know, cos A is defined as follows
Now we find the value of expression sin A – cos A
Therefore by substituting the value the value of sin A and cos A from equation (4) and (5) respectively, we get,
If tan θ = 20/21, show that
Given: tan θ = 20/21
tan θ = 20/21
Side AC be the hypotenuse and can be found by applying Pythagoras theorem
AC^{2}= 21^{2} + 20^{2}
AC^{2} = 441 + 400
AC^{2} = 841
AC = √841
AC = 29
Therefore Hypotenuse side AC = 29
Now we know, sin θ is defined as follows,
Therefore from figure and above equation
Now we need to find the value of expression
Therefore by substituting the value of sin θ and cos θ from above equations, we get
Therefore after evaluating we get,
If cosec A = 2, find
Given: cosec A = 2
Here BC is the adjacent side,
By applying Pythagoras theorem,
4 = 1 + BC^{2}
BC^{2 }= 3
BC = √3
Substitute all the values of sin A, cos A and tan A from the equations (1), (2) and (3) respectively
= 2
If ∠A and ∠B are acute angles such that cos A =cos B, then show that ∠A = ∠B
∠A and ∠B are acute angles
cos A = cos B such that ∠A = ∠B
Let us consider right angled triangle ACB
Now since cos A = cos B
Now observe that denominator of above equality is same that is AB
Therefore AC = BC
We know that when two sides of triangle are equal, then opposite of the sides are also
Equal.
We can say that
Angle opposite to side AC = angle opposite to side BC
∠B = ∠A
Hence, ∠A = ∠B
In a ΔABC, right angled triangle at A, if tan C = √3, find the value of sin B cos C + cos B sin C.
Given: ΔABC
To find: sin B cos C + cos B sin C
The given a ΔABC is as shown in figure
Side BC is unknown and can be found using Pythagoras theorem,
BC^{2 }= AB^{2} + AC^{2}
BC^{2 }= 3 +1
BC^{2} = 4
BC = √4
BC = 2
Therefore Hypotenuse side BC = 2 … (1)
Now by substituting the values from equation (1) and figure
Now substituting the value from equation
Similarly
So by evaluating
Now, by substituting the value of sin B, cos B, sin C and cos C from equation (2), (3), (4) and (5) respectively in sin B cos C + cos B sin C
Sin B cos C + cos B sin C
= 1
sin B cos C + cos B sin C = 1
State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of ∠A.
(iii) cos A is the abbreviation used for the cosecant of ∠A.
(iv) sin θ = 4/3 for some angle θ.
(i) tan A < 1
Value of tan A at 45^{°} i.e… tan 45 = 1
As value as A increases to 90^{°}
Tan A becomes infinite
So given statement is false.
(ii) sec A = 12/5 for some value of angle if
M-I
sec A = 2.4
sec A > 1
So given statements is true.
M- II
For sec A = 12/5 we get adjacent side = 13
Subtending 9i at B.
So, given statement is true.
(iii) Cos A is the abbreviation used for cosecant of angle A.
The given statement is false.
As such cos A is the abbreviation used for cos of angle A, not as cosecant of angle A.
(iv) cot A is the product of cot A and A
Given statement is false? cot A is a co-tangent of angle A and co-tangent of angle
(v) sin θ = 4/3 for some angle θ.
Given statement is false
Since value of sin θ is less than (or) equal to one.
Here value of sin θ exceeds one,
As shown in figure
169 = 144 + BC^{2}
BC^{2 }= 169 – 144
BC^{2 }= 25
BC = 5
Now we know that,
We also know that,
Therefore, substituting the value of sin θ and cos θ from above equations
Now substitute all the values of sin θ, cos θ and tan θ from above equations in
Therefore by further simplifying we get,
If cos θ = 5/13, find the value of
Given: If cos θ = 5/13 ….. (1)
To find: The value of expression
Now when we compare equation (1) and (2)
Therefore, Triangle representing ∠θ is as shown figure.
Perpendicular side AB is unknown and it can be found by using Pythagoras theorem
Therefore by substituting the values of known sides,
AB^{2} = 13^{2} – 5^{2}
AB^{2}= 169 -25
Now we know from figure and equation,
Now w substitute all the values from equation (1), (4) and (5) in the expression below,
Given: sec A = 17/8
Now, by substituting the value of sec A
Now we also know that,
sin^{2}A + cos^{2}A = 1
sin^{2}A = 1 – cos^{2}A
sin A = 15/17
Now by substituting the value of all the terms,
tan A = 15/8
Now from the expression of above equation which we want to prove:
Now by substituting the value of cos A ad sin A from equation (3) and (4)
From expression
Now by substituting the value of tan A from above equation
We can see that,
If sin θ = 3/4,
Prove that
To prove:
By comparing (1) and (3)
Perpendicular side = 3 and
Hypotenuse = 4
Side BC is unknown.
So we find BC by applying Pythagoras theorem to right angled ΔABC
Now we substitute the value of perpendicular side (AB) and hypotenuse (AC) and get the base side (BC)
4^{2 }= 3^{2} + BC^{2}
BC^{2} = 16 – 9
BC^{2 }= 7
BC = √7
Hence, Base side BC = √7 … (3)
Now cos A = BC/AC
√7/4 … (4)
Therefore, from fig and equation (1)
cosec A = 4/3 … (5)
Now, similarly
Sec A = 4/√7 … (6)
Further we also know that
Therefore by substituting the values from equation (1) and (4),
Now by substituting the value of cosec A, sec A and cot A from the equations (5), (6), and (7) in the L.H.S of expression (2)
Hence it is proved that,
If sec A = 17/8,
Verify that
Given: sec A = 17/8 … (1)
To verify:
sec A = 1/(cos A)
cos A = 1/(sec A)
cos A = 8/17 … (3)
Similarly we can also get,
sin A = 15/17 …. (4)
An also we know that
tan A = (sin A)/(cos A)
tan A = 15/8 … (5)
Now from the expression of equation (2)
L.H.S: Missing close
Now by substituting the value of cos A and sin A from equation (3) and (4)
Now by substituting the value of tan A from equation (5)
Now by comparing equation (6) and (7)
If cot θ = 3/4, prove that
Given: cot θ = 3/4
Prove that:
Here AC is the hypotenuse and we can find that by applying Pythagoras theorem
AC^{2 }= AB^{2} +BC^{2}
AC^{2} = 16 +9
AC^{2 }= 25
Now on substituting the values in equations we get,
If 3 cos θ – 4 sin θ = 2 cos θ + sin θ, find tan θ tan θ
Given: 3 cos θ - 4 sin θ = 2 cos θ + sin θ
To find: tan θ tan θ
We can write this as:
3 cos θ – 4 sin θ = 2 cos θ + sin θ cos θ = 5 sin θ
Dividing both the sides by cos θ,
1 = 5 tan θ
tan θ = 1
If ∠A and ∠P are acute angles such that tan A = tan P, then show ∠A = ∠P
Given: A and P are acute angles tan A = tan P
Prove that: ∠A = ∠P
Let us consider right angled triangle ACP
We know
∴ tan A = tan P
PC = AC [∵ Angle opposite to equal sides are equal]
∠A = ∠P
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Chapter 5: Trigonometric Ratios Exercise –...