Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
Chapter 6: Trigonometric Identities Exercise – 6.1 Question: 1 (1 - cos2 A) cosec2 A = 1 Solution: (1 - cos2 A)cosec2 A = Sin2 A cosec2 A = (Sin A cosec A)2 = (Sin A × (1/Sin A))2 = (1)2 = 1 Question: 2 (1 + Cot2 A) Sin2 A = 1 Solution: We know, cosec2A - Cot 2 A = 1 So, (1 + Cot2 A) Sin2 A = Cosec2 A Sin2 A = (Cosec A Sin A)2 = ((1/Sin A) × Sin A)2 = (1)2 = 1 Question: 3 tan2θ cos2θ = 1 − cos2θ Solution: We know, sin2 θ + cos2 θ = 1 So, tan2 θ cos2 θ = (tan θ × cos θ)2 = (sin θ)2 sin2θ 1 - cos2θ Question: 4 Solution: We know, sin2 θ + cos2 θ = 1 = 1 Question: 5 (sec2θ − 1)(cosec2θ − 1) = 1 Solution: We know that, (sec2θ − tan2θ) = 1 (cosec2θ − cot2θ) = 1 So, (sec2θ - 1)(cosec2θ - 1) = tan2θ × cot2θ = (tan θ × cot θ)2 = 12 = 1 Question: 6 Solution: We know that, (sec2θ − tan2θ) = 1 So, Question: 7 Solution: We know, sin2θ + cos2θ = 1 So, Multiplying both numerator and denominator by (1+ sin θ), we have Question: 8 Solution: We know, sin2θ + cos2θ = 1 Multiplying both numerator and denominator by (1− sin θ), we have Question: 9 Solution: We know that, sin2A + cos2A = 1 cosec2A - cot2A = 1 = cos2A + sin2A =1 Question: 10 Solution: We know, sin2A + cos2A = 1 sec2A – tan2A = 1 = 1 Question: 11 Solution: Question: 12 Solution: We know, sin2θ + cos2θ = 1 Multiplying both numerator and denominator by (1+ cos θ), we have Question: 13 Solution: We know that, (a - b)(a + b) = a2 - b2 Here, (1 - cos2θ) = sin2 θ ⟹ cosec θ + cot θ Hence, L.H.S = R.H.S Question: 14 Solution: Given, Rationalize with nr and dr with 1 – sin θ Here, (1 - sin θ)(1 + sin θ) = cos2θ ⟹ (sec θ – tan θ) 2 Hence, L.H.S = R.H.S Question: 15 Solution: Given, Here, 1 + cot2 θ = cosec2 θ ⟹ cot θ Hence, L. H. S = R.H.S Question: 16 tan2θ − sin2θ = tan2θ ∗ sin2θ Solution: Given, L.H.S = tan2θ − sin2θ ⟹ tan2θ *sin2θ Hence, L.H.S = R.H.S Question: 17 (cosec θ + sin θ)(cosec θ - sin θ) = cot2θ + cos2θ Solution: Given, L.H.S = (cosec θ + sin θ)(cosec θ - sin θ) Here, (a + b)(a - b) = a2 - b2 cosec2 θ can be written as 1 + cot2 θ and sin2 θ can be written as 1 - cos2 θ ⟹ 1 + cot2 θ - (1 - cos2 θ) ⟹ 1 + cot2 θ - 1 + cos2 θ ⟹ cot2 θ + cos2 θ Hence, L.H.S = R.H.S Question: 18 (sec θ + cos θ) (sec θ - cos θ) = tan2θ + sin2θ Solution: Given, L.H.S = (sec θ + cos θ)(sec θ - cos θ) Here, (a + b)(a - b) = a2 - b2 sec2θ can be written as 1 + tan2 θ and cos2 θ can be written as 1 - sin2 θ ⟹ 1 + tan2 θ - (1 - sin2 θ) ⟹ 1 + tan2 θ - 1 + sin2 θ ⟹ tan2θ + sin2 θ Hence, L.H.S = R.H.S Question: 19 Sec A(1- sin A) (sec A + tan A) = 1 Solution: Given, L.H.S = sec A(1 – sin A)(sec A + tan A) ⟹ 1 Hence, L.H.S = R.H.S Question: 20 (cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1 Solution: Given, L.H.S = sec A(1 – sin A)(sec A + tan A) Substitute the above values in L.H.S ⟹ 1 Hence, L. H. S = R. H. S Question: 21 (1 + tan2θ)(1 - sin θ)(1 + sin θ) = 1 Solution: Given, L.H.S = (1 + tan2θ)(1 - sin θ)(1 + sin θ) We know that, Sin2 θ + cos2 θ = 1 and sec2 θ - tan2 θ = 1 So, (1 + tan2θ)(1 - sin θ)(1 + sin θ) = (1 + tan2θ){(1 - sin θ)(1 + sin θ)} = (1 + tan2θ)(1 - sin2θ) Hence, L.H.S = R.H.S Question: 22 (sin2A ∗ cot2A) + (cos2A; ∗ tan2A) = 1 Solution: Given, = cos2A + sin2A = 1 Hence, L.H.S = R.H.S Question: 23 Solution: 1. Give, L.H.S = cot θ - tan θ Here, Sin2 θ + cos2θ = 1 So, ⟹ cot θ - tan θ Hence, L.H.S = R.H.S Given, L. H. S = tan θ – cot θ We know that, Sin2θ + cos2θ = 1 tan θ – cot θ Hence, L. H. S = R. H. S Question: 24 Solution: Given, L.H.S We know that, Sin2θ + cos2 θ = 1 = - sin θ + sin θ = 0 Hence, L.H.S = R.H.S Question: 25 Solution: ⇒ 2sec2A LHS = RHS Hence proved Question: 26 Solution: = 2 sec θ LHS = RHS Hence proved Question: 27 Solution: We know that sin2θ + cos2θ = 1 LHS = RHS Hence prove Question: 28 Solution: LHS = RHS Hence proved Question: 29 Solution: = 1 + cos θ LHS = RHS Hence proved Question: 30 Solution: = 1 + tan θ + cot θ LHS = RHS Hence proved Question: 31 sec6θ = tan6θ + 3 tan2 θ sec2θ + 1 Solution: We know that sec2θ − tan2θ = 1 Cubing both sides (sec2θ − tan2θ)3= 1 sec6θ − tan6θ − 3sec2θ tan2θ(sec2θ − tan2θ) = 1 [Since, a3 − b3 = (a − b)(a2 + ab + b2)] sec6θ − tan6θ − 3sec2θ tan2θ = 1 ⇒ sec6θ = tan6θ + 3sec2θ tan2θ + 1 Hence proved. Question: 32 cosec6θ = cot6θ + 3cot2θ cosec2θ + 1 Solution: We know that cosec2θ − cot2θ = 1 Cubing both sides (cosec2θ − cot2θ)3 = 1 cosec6θ − cot6θ − 3cosec2θ cot2θ (cosec2θ − cot2θ) = 1 [Since, a3 − b3 = (a − b)(a2 + ab + b2)] cosec6θ − cot6θ − 3cosec2θ cot2θ = 1 ⇒ cosec6θ = cot6θ + 3 cosec2θ cot2θ + 1 Hence proved. Question: 33 Solution: We know that sec2θ − tan2θ = 1 Therefore, sec2θ = 1 + tan2θ LHS = RHS Hence proved Question: 34 Solution: We know that sin2A + cos2A = 1 sin2A = 1 − cos2A ⇒ sin2A = (1 – cos A)(1 + cos A) LHS = RHS Hence proved Question: 35 Solution: Rationalizing the denominator by multiplying and dividing with Sec A + tan A, we get LHS = RHS Hence proved Question: 36 Solution: Multiply both numerator and denominator with (1 – cos A) LHS = RHS Hence proved Question: 37 Solution: Considering left hand side (L. H. S), Rationalize the numerator and denominator with Therefore, LHS = RHS Hence proved Considering left hand side (LHS), Rationalize the numerator and denominator. Therefore, LHS = RHS Hence proved Question: 38 Prove that: Solution: Therefore, LHS = RHS Hence proved Therefore, LHS = RHS Hence proved Considering left hand side (LHS), Rationalize the numerator and denominator. Therefore, LHS = RHS Hence proved Considering left hand side (LHS), Multiply and divide with (1 + cos θ) Therefore, LHS = RHS Hence proved Question: 39 Solution: Considering left hand side (LHS), = (sec A – tan A)2 Therefore, LHS = RHS Hence proved Question: 40 Solution: Considering left hand side (LHS), Rationalize the numerator and denominator with (1 - cos A) = (cosec A - cot A)2 = (cot A - cosec)2 Therefore, LHS = RHS Hence proved Question: 41 Solution: Considering left hand side (LHS), = 2cosec A cot A Therefore, LHS = RHS Hence proved Question: 42 Solution: Considering left hand side (LHS), = cos A + sin A Therefore, LHS = RHS Hence proved Question: 43 Solution: Considering left hand side (LHS), = 2 sec2 A Therefore, LHS = RHS Hence proved Question: 44 Solution: Considering left hand side (LHS), Therefore, LHS = RHS Hence proved Question: 45 Solution: Considering left hand side (LHS), Therefore, LHS = RHS Hence proved Question: 46 Solution: Considering left hand side (LHS), = cot θ Therefore, LHS = RHS Hence, proved. Question: 47 Solution: Dividing the numerator and denominator with cos θ Considering LHS, we get, Therefore, LHS = RHS Hence proved Dividing the numerator and denominator with cos θ, we get, Therefore, LHS = RHS Hence proved Question: 48 Solution: Considering LHS, we get, dividing the numerator and denominator with sin θ, we get, = cosec θ + cot θ Therefore, LHS = RHS Hence proved Question: 49 (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ Solution: To prove, (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ Considering LHS, we get, = sec θ + cosec θ Therefore, LHS = RHS Hence proved Question: 50 Solution: To prove, Considering LHS, we get, = 2 cosec A Therefore, LHS = RHS Hence proved Question: 51 Solution: Considering LHS, we get, = 1 + cosec θ - 1 [∴ (a + b)(a - b) = a2- b2] = cosec θ [ a = cosec θ, b = 1] Therefore, LHS = RHS Hence, proved Question: 52 Solution: Therefore, LHS = RHS Hence, proved Question: 53 Solution: Therefore, LHS = RHS. Hence Proved. Question: 54 sin2A cos2B - cos2A sin2B = sin2A - sin2B Solution: LHS = sin2Acos2B - cos2A sin2B = sin2A− sin2A sin2 B − sin2 B + sin2A sin2 B = sin2A − sin2 B RHS Hence Proved. Question: 55 Solution: = cot A tan B = RHS Hence Proved. = tan A tan B = RHS Hence Proved. Question: 56 cot2A cosec2B − cot2B cosec2A = cot2A − cot2B Solution: LHS = cot2A cosec2B − cot2 B cosec2A = cot2A + cot2A cot2B − cot2B − cot2B cot2A = cot2A - cot2B RHS Hence Proved. Question: 57 tan2A sec2B − sec2A tan2B = tan2A − tan2B Solution: LHS = tan2A sec2B − sec2A tan2B = tan2A(1 + tan2B) − sec2A(tan2A) = tan2A + tan2A tan2B − tan2B − tan2A tan2B = tan2A − tan2B = RHS Hence Proved. Question: 58 If x = a sec θ + b tan θ and y = a tan θ + b sec θ, Prove that x2 - y2 = a2 - b2. Solution: LHS = x2 - y2 = (a sec θ + b tan θ)2 − (a tan θ + b sec θ)2 = a2sec2θ + b2tan2θ + 2 a b sec θ tan θ − a2 tan2θ − b2sec2θ – 2 ab sec θ tan θ = a2sec2θ + b2tan2θ − a2tan2θ − b2sec2θ = a2sec2θ − b2sec2θ + b2tan2θ − a2tan2θ = sec2θ (a2 − b2) + tan2θ (b2 − a2) = sec2θ (a2 − b2) − tan2θ (a2 − b2) = (sec2θ − tan2θ) (a2 − b2) = a2 - b2 RHS Hence Proved. Question: 59 If 3 sin θ + 5 cos θ = 5, Prove that 5 sin θ – 3 cos θ = ± 3. Solution: Given 3 sin θ + 5 cos θ = 5 3 sin θ = 5 – 5 cos θ 3 sin θ = 5(1 – cos θ) 3+ 3 cos θ = 5sin θ 5 sin θ - 3 cos θ = ±3 = RHS Hence Proved. 5 sin θ – 3 cos θ = ± 3. Question: 60 If cosec θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1. Solution: LHS = m n = (cosec θ + cot θ) (cosec θ – cot θ) = cosec2θ − cot2θ = 1 = RHS Hence Proved. Question: 61 If Tn = sinn θ + cosn θ, Prove that Solution: = sin2θ cos2θ LHS = RHS Hence proved. Question: 62 Solution: (tan θ + sec θ)2 + (tan θ – sec θ)2 = tan2θ + sec2θ + 2 tan θ sec θ + tan2 θ + sec2θ – 2 tan θ sec θ = 2 tan 2θ + 2 sec 2θ = 2[tan2θ + sec2θ] LHS = RHS Hence proved. Question: 63 Solution: LHS = RHS Hence proved. Question: 64 Solution: LHS = RHS Hence proved. LHS = RHS Hence proved. Question: 65 (sec A + tan A - 1) (sec A - tan A + 1) = 2 tan A Solution: = (sec A + tan A − {sec2A - tan2A}) [sec A – tan A + (sec2A − tan2A)] = (sec A + tan A − (sec A + tan A) (sec A – tan A)) [sec A – tan A + (sec A – tan A)(sec A + tan A)] = (sec A + tan A) (1 − (sec A – tan A)) (sec A – tan A) (1 + (sec A + tan A)) = (sec2A − tan2A) (1 – sec A + tan A) (1 + sec A + tan A) =2 tan A LHS = RHS Hence proved. Question: 66 (1 + cot A - cosec A)(1 + tan A + sec A) = 2 Solution: LHS = (1 + cot A - cosec A) (1 + tan A + sec A) LHS = RHS Hence proved. Question: 67 (cosec θ – sec θ) (cot θ – tan θ) = (cosec θ + sec θ) (sec θ cosec θ − 2) Solution: LHS = (cosec θ – sec θ) (cot θ – tan θ) RHS = (cosec θ + sec θ) (sec θ cosec θ - 2) LHS = RHS Hence proved. Question: 68 Solution: = cosec A - sec A = RHS = LHS Hence proved. Question: 69 Solution: = LHS = RHS Hence proved. Question: 70 Solution: = sin A × cos3 A +cos A × sin3A = sin A cos A (cos2A + sin2A) = sin A cos A LHS = RHS Hence proved. Question: 71 sec4A (1 − sin4A) – 2 tan2A = 1 Solution: Given, L.H.S = sec4A (1 - sin4A) - 2tan2A = sec4A − sec4A × sin4A − 2tan4A = sec4A - tan4A – 2 tan4A = (sec2A)2 - tan4A – 2 tan4A = (1+ tan2A)2 − tan4A − 2tan4A = 1 + tan4A + 2tan2A − tan4A − 2tan4A = 1 Hence, L.H.S = R.H.S Question: 72 Solution: Here, sin2A + cos2A = 1 Solving, RHS ⟹ Multiplying Nr. And Dr. with (1 + Sin A) Hence, LHS = RHS Question: 73 Solution: Given, L.H.S = (1 + cot A + tan A) (sin A – cos A) ⟹ sin A – cos A + cot A sin A – cot A cos A + sin A tan A – tan A cos A ⟹ sin A tan A – cos A cot A Hence, L.H.S = R.H.S Question: 74 Solution: Question: 75 If cosec θ – sin θ = a3, sec θ – cos θ = b3, Prove that a2b2 (a2+ b2) = 1 Solution: Given, cosec θ – sin θ = a3 Here cos2θ A + sin2 A = 1 Squaring on both sides Squaring on both sides = 1 Hence, L.H.S = R.H.S Question: 76 If a cos3 θ + 3 a cos θ sin2θ = m, a sin3θ + 3 a cos2θ sin θ = n, Solution: Substitute the values of m and n in the above equation ⟹ a cos3θ + 3 a cos θ sin2 θ + a sin3 θ + 3 a cos2θ sin2/3 θ + a cos3 θ + 3 a cos θ sin2 θ – a sin3 θ – 3 a cos2 θ sin2/3 θ ⟹ (a)2/3 cos3θ + 3 cos θ sin2 θ + sin3θ + 3 cos2 θ sin2/3θ + (a)2/3 cos3 θ + 3 cos θ sin 2θ - sin3 θ – 3 cos2 θ sin2/3θ ⟹ (a)2/3((cos θ + sin θ)3)2/3 + (a)2/3 ((cos θ – sin θ)3)2/3 ⟹ (a)2/3[(cos θ + sin θ)2] + (a)2/3[(cos θ – sin θ)2] ⟹ (a)2/3((cos2θ + sin2θ + 2 sin θ cos θ)) + (a)2/3((cos2θ + sin2θ – 2 sin θ cos θ)) ⟹ (a)2/3[ 1 + 2 sin θ cos θ] + (a)2/3[1 – 2 sin θ cos θ] ⟹ (a)2/3[1 + 2 sin θ cos θ] + 1 – 2 sin θ cos θ] ⟹ (a)2/3(1 + 1) ⟹ 2(a)2/3 Hence, L.H.S = R.H.S Question: 77 If x = a cos3θ, y = b sin3θ, Solution: x = a cos3 θ: y = b sin3 θ = 1 Hence proved. Question: 78 If a cos θ + b sin θ = m and a sin θ – b cos θ = n, Prove that a2 + b2 = m2 + n2 Solution: R.H.S = m2 + n2 = m2 + n2 = (a cos θ + b sin θ)2 + (a sin θ – b cos θ)2 = a2 cos2 θ + b2 sin2θ + 2ab sin θ cos θ + a2 + sin2θ + b2 cos2θ – 2ab sin θ cos θ = a2 cos2θ + b2 cos2θ + b2 sin2θ + a2 sin2θ = a2(sin2 θ + cos2 θ) + b2(sin2 θ + cos2 θ) = a2 + b2 (∵ sin2 θ + cos2θ = 1) Question: 79 If cos A + cos2A = 1, Prove that sin2A + sin4A = 1 Solution: Given - cos A + cos2 A = 1 We have to prove sin2 A + sin4 A = 1 Now, cos A + cos2 A = 1 cos A = 1 - cos2 A cos A = sin2 A sin2 A = cos A Therefore, We have sin2 A + sin4 A = cos A + (cos A)2 = cos A + cos2 A = 1 Hence proved. Question: 80 If cos θ + cos2 θ = 1, Prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1 Solution: cos θ + cos2 θ = 1 cos θ = 1 − cos2 θ cos θ = sin2 θ (i) Now, sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = (sin4 θ)3 + 3 sin4 θ. sin2 θ (sin4 θ + sin2 θ) + (sin2 θ)3 + 2(sin2 θ)2 + 2 sin2 θ − 2 Using (a + b)3 = a3 + b3 + 3 ab(a + b) and also from(i) cos θ = sin2 θ (sin4 θ + sin2 θ)3 + 2cos2 θ + 2 cos θ − 2 ((sin2 θ)2 + sin2 θ)3 + 2 cos2 θ + 2 cos θ – 2 (cos2 θ + sin2 θ)3 + 2 cos2 θ + 2 cos θ − 2 1 + 2(cos2 θ + sin2 θ) − 2 = 1 + 2(1) −2 = 1 L.H.S = R.H.S Hence proved. Question: 81 Given that: (1 + cos α)(1 + cos β)(1 + cos γ) = (1 – cos α)(1 – cos β)(1 – cos γ). Show that one of the values of each member of this equality is sin α sin β sin γ. Solution: We know that Multiply (i) with sin α sin β sin γ and divide it with same we get Hence sin α sin β sin γ is the member of equality. Question: 82 If sin θ + cos θ = x, Solution: sin θ + cos θ = x Squaring on both sides (sin θ + cos θ)2 = x2 ⇒ sin2θ + cos2θ + 2 sin θ cos θ = x2 We know that sin2 θ + cos2 θ = 1 Cubing on both sides (sin2 θ + cos2 θ)3 = 13 sin6 θ + cos6 θ + 3 sin2 θ cos2 θ (sin2 θ + cos2 θ) = 1 ⇒ sin6 θ + cos6 θ = 1 – 3 sin2 θ cos2θ Question: 83 If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan ϕ, Solution: Given, x = a sec θ cos ϕ, y = b sec θ sin ϕ, z = c tan ϕ squaring x, y, z on the sides x2 = a2sec2θ cos2 ϕ x2/a2 = sec2θ cos2 ϕ --- 1 y2 = b2 sec2θ sin2 ϕ y2/b2 = sec2θ sin2 ϕ --- 2 z2 = c2 tan2 ϕ z2/c2 = tan2 ϕ --- 3 Substitute equation 1, 2, 3 in x2/a2 + y2/b2 − z2/c2 ⟹ x2/a2 + y2/b2 − z2/c2 ⟹ Sec2 θ cos2 ϕ + sec2 θ sin2 ϕ - tan2 ϕ ⟹ sec2θ(cos2 ϕ + sin2 ϕ) - tan2 ϕ We know that, cos2 ϕ + sin2 ϕ = 1 ⟹ sec2θ (1) - tan2 ϕ and sec2θ − tan2θ = 1 ⟹ 1 Hence, L.H.S= R.H.S Question: 84 If sin θ + 2 cos θ. Prove that 2 sin θ – cos θ = 2 Solution: Given, sin θ + 2 cos θ = 1 Squaring on both sides ⟹ (sin θ + 2 cos θ)2 = 12 ⟹ sin2θ + 4 cos2θ + 4 sin θ cos θ = 1 ⟹ 4 cos2 θ + 4 sin θ cos θ = 1 - sin2θ Here, 1 - sin2θ = cos2θ ⟹ 4 cos2θ + 4 sin θ cos θ - cos2θ = 0 ⟹ 3 cos2θ + 4 sin θ cos θ = 0 ---- 1 We have, 2 sin θ – cos θ = 2 Squaring L.H.S (2 sin θ – cos θ)2 = 4 sin2θ + cos2θ – 4 sin θ cos θ Here, 4 sin θ cos θ = 3 cos2θ = 4 sin2θ + cos2θ + 3 cos2θ = 4 sin2 θ + 4 cos2 θ = 4(sin2θ + cos2θ) = 4(1) = 4 (2 sin θ – cos θ)2 = 4 ⟹ 2sin θ – cos θ = 2 Hence proved
(1 - cos2 A) cosec2 A = 1
(1 - cos2 A)cosec2 A
= Sin2 A cosec2 A
= (Sin A cosec A)2
= (Sin A × (1/Sin A))2
= (1)2
= 1
(1 + Cot2 A) Sin2 A = 1
We know,
cosec2A - Cot 2 A = 1
So, (1 + Cot2 A) Sin2 A
= Cosec2 A Sin2 A
= (Cosec A Sin A)2
= ((1/Sin A) × Sin A)2
tan2θ cos2θ = 1 − cos2θ
sin2 θ + cos2 θ = 1
So, tan2 θ cos2 θ
= (tan θ × cos θ)2
= (sin θ)2 sin2θ 1 - cos2θ
(sec2θ − 1)(cosec2θ − 1) = 1
We know that, (sec2θ − tan2θ) = 1 (cosec2θ − cot2θ) = 1
So, (sec2θ - 1)(cosec2θ - 1)
= tan2θ × cot2θ
= (tan θ × cot θ)2
= 12
We know that, (sec2θ − tan2θ) = 1
So,
We know, sin2θ + cos2θ = 1
So, Multiplying both numerator and denominator by (1+ sin θ), we have
We know, sin2θ + cos2θ = 1 Multiplying both numerator and denominator by (1− sin θ), we have
We know that, sin2A + cos2A = 1 cosec2A - cot2A = 1
= cos2A + sin2A
=1
We know, sin2A + cos2A = 1 sec2A – tan2A = 1
We know, sin2θ + cos2θ = 1 Multiplying both numerator and denominator by (1+ cos θ), we have
We know that, (a - b)(a + b) = a2 - b2
Here, (1 - cos2θ) = sin2 θ
⟹ cosec θ + cot θ
Hence, L.H.S = R.H.S
Given,
Rationalize with nr and dr with 1 – sin θ
Here, (1 - sin θ)(1 + sin θ) = cos2θ
⟹ (sec θ – tan θ) 2
Here, 1 + cot2 θ = cosec2 θ
⟹ cot θ
Hence, L. H. S = R.H.S
tan2θ − sin2θ = tan2θ ∗ sin2θ
Given, L.H.S = tan2θ − sin2θ
⟹ tan2θ *sin2θ
(cosec θ + sin θ)(cosec θ - sin θ) = cot2θ + cos2θ
Given, L.H.S = (cosec θ + sin θ)(cosec θ - sin θ)
Here, (a + b)(a - b) = a2 - b2 cosec2 θ can be written as 1 + cot2 θ and sin2 θ can be written
as 1 - cos2 θ
⟹ 1 + cot2 θ - (1 - cos2 θ)
⟹ 1 + cot2 θ - 1 + cos2 θ
⟹ cot2 θ + cos2 θ
(sec θ + cos θ) (sec θ - cos θ) = tan2θ + sin2θ
Given, L.H.S = (sec θ + cos θ)(sec θ - cos θ)
Here, (a + b)(a - b) = a2 - b2 sec2θ can be written as 1 + tan2 θ and cos2 θ can be written as 1 - sin2 θ
⟹ 1 + tan2 θ - (1 - sin2 θ)
⟹ 1 + tan2 θ - 1 + sin2 θ
⟹ tan2θ + sin2 θ
Sec A(1- sin A) (sec A + tan A) = 1
Given, L.H.S = sec A(1 – sin A)(sec A + tan A)
⟹ 1
(cosec A – sin A)(sec A – cos A)(tan A + cot A) = 1
Substitute the above values in L.H.S
Hence, L. H. S = R. H. S
(1 + tan2θ)(1 - sin θ)(1 + sin θ) = 1
Given, L.H.S = (1 + tan2θ)(1 - sin θ)(1 + sin θ)
We know that, Sin2 θ + cos2 θ = 1 and sec2 θ - tan2 θ = 1
So, (1 + tan2θ)(1 - sin θ)(1 + sin θ)
= (1 + tan2θ){(1 - sin θ)(1 + sin θ)}
= (1 + tan2θ)(1 - sin2θ)
(sin2A ∗ cot2A) + (cos2A; ∗ tan2A) = 1
= cos2A + sin2A = 1
1. Give, L.H.S = cot θ - tan θ
Here, Sin2 θ + cos2θ = 1
So, ⟹ cot θ - tan θ
Given, L. H. S = tan θ – cot θ
We know that,
Sin2θ + cos2θ = 1
tan θ – cot θ
Given, L.H.S
Sin2θ + cos2 θ = 1
= - sin θ + sin θ = 0
⇒ 2sec2A
LHS = RHS
Hence proved
= 2 sec θ
We know that sin2θ + cos2θ = 1
Hence prove
LHS = RHS Hence proved
= 1 + cos θ
= 1 + tan θ + cot θ
sec6θ = tan6θ + 3 tan2 θ sec2θ + 1
We know that sec2θ − tan2θ = 1
Cubing both sides
(sec2θ − tan2θ)3= 1 sec6θ − tan6θ − 3sec2θ tan2θ(sec2θ − tan2θ) = 1 [Since, a3 − b3 = (a − b)(a2 + ab + b2)]
sec6θ − tan6θ − 3sec2θ tan2θ = 1
⇒ sec6θ = tan6θ + 3sec2θ tan2θ + 1
Hence proved.
cosec6θ = cot6θ + 3cot2θ cosec2θ + 1
We know that cosec2θ − cot2θ = 1
Cubing both sides (cosec2θ − cot2θ)3 = 1
cosec6θ − cot6θ − 3cosec2θ cot2θ (cosec2θ − cot2θ) = 1 [Since, a3 − b3 = (a − b)(a2 + ab + b2)]
cosec6θ − cot6θ − 3cosec2θ cot2θ = 1
⇒ cosec6θ = cot6θ + 3 cosec2θ cot2θ + 1
Therefore, sec2θ = 1 + tan2θ
We know that
sin2A + cos2A = 1 sin2A = 1 − cos2A
⇒ sin2A = (1 – cos A)(1 + cos A)
Rationalizing the denominator by multiplying and dividing with
Sec A + tan A, we get
Multiply both numerator and denominator with (1 – cos A)
Considering left hand side (L. H. S), Rationalize the numerator and denominator with
Therefore,
Considering left hand side (LHS), Rationalize the numerator and denominator.
Therefore, LHS = RHS
Prove that:
Considering left hand side (LHS),
Multiply and divide with (1 + cos θ)
Considering left hand side (LHS), = (sec A – tan A)2
Therefore, LHS = RHS Hence proved
Considering left hand side (LHS), Rationalize the numerator and denominator with (1 - cos A)
= (cosec A - cot A)2 = (cot A - cosec)2
= 2cosec A cot A
= cos A + sin A
= 2 sec2 A
= cot θ
Hence, proved.
Dividing the numerator and denominator with cos θ Considering LHS, we get,
Dividing the numerator and denominator with cos θ, we get,
Considering LHS, we get, dividing the numerator and denominator with sin θ, we get,
= cosec θ + cot θ
(sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ
To prove, (sin θ + cos θ) (tan θ + cot θ) = sec θ + cosec θ
Considering LHS, we get,
= sec θ + cosec θ
To prove,
= 2 cosec A
= 1 + cosec θ - 1 [∴ (a + b)(a - b) = a2- b2]
= cosec θ [ a = cosec θ, b = 1]
Therefore, LHS = RHS Hence, proved
Therefore, LHS = RHS. Hence Proved.
sin2A cos2B - cos2A sin2B = sin2A - sin2B
LHS = sin2Acos2B - cos2A sin2B
= sin2A− sin2A sin2 B − sin2 B + sin2A sin2 B
= sin2A − sin2 B
RHS Hence Proved.
= cot A tan B
= RHS Hence Proved.
= tan A tan B
cot2A cosec2B − cot2B cosec2A = cot2A − cot2B
LHS = cot2A cosec2B − cot2 B cosec2A
= cot2A + cot2A cot2B − cot2B − cot2B cot2A
= cot2A - cot2B
tan2A sec2B − sec2A tan2B = tan2A − tan2B
LHS = tan2A sec2B − sec2A tan2B
= tan2A(1 + tan2B) − sec2A(tan2A)
= tan2A + tan2A tan2B − tan2B − tan2A tan2B
= tan2A − tan2B
If x = a sec θ + b tan θ and y = a tan θ + b sec θ, Prove that x2 - y2 = a2 - b2.
LHS = x2 - y2 = (a sec θ + b tan θ)2 − (a tan θ + b sec θ)2
= a2sec2θ + b2tan2θ + 2 a b sec θ tan θ − a2 tan2θ − b2sec2θ – 2 ab sec θ tan θ
= a2sec2θ + b2tan2θ − a2tan2θ − b2sec2θ
= a2sec2θ − b2sec2θ + b2tan2θ − a2tan2θ
= sec2θ (a2 − b2) + tan2θ (b2 − a2)
= sec2θ (a2 − b2) − tan2θ (a2 − b2)
= (sec2θ − tan2θ) (a2 − b2) = a2 - b2
If 3 sin θ + 5 cos θ = 5,
Prove that 5 sin θ – 3 cos θ = ± 3.
Given 3 sin θ + 5 cos θ = 5
3 sin θ = 5 – 5 cos θ
3 sin θ = 5(1 – cos θ)
3+ 3 cos θ = 5sin θ
5 sin θ - 3 cos θ = ±3
= RHS Hence Proved. 5 sin θ – 3 cos θ = ± 3.
If cosec θ + cot θ = m and cosec θ – cot θ = n, prove that mn = 1.
LHS = m n = (cosec θ + cot θ) (cosec θ – cot θ)
= cosec2θ − cot2θ
If Tn = sinn θ + cosn θ, Prove that
= sin2θ cos2θ
(tan θ + sec θ)2 + (tan θ – sec θ)2 = tan2θ + sec2θ + 2 tan θ sec θ + tan2 θ + sec2θ – 2 tan θ sec θ
= 2 tan 2θ + 2 sec 2θ
= 2[tan2θ + sec2θ]
LHS = RHS Hence proved.
(sec A + tan A - 1) (sec A - tan A + 1) = 2 tan A
= (sec A + tan A − {sec2A - tan2A}) [sec A – tan A + (sec2A − tan2A)]
= (sec A + tan A − (sec A + tan A) (sec A – tan A)) [sec A – tan A + (sec A – tan A)(sec A + tan A)]
= (sec A + tan A) (1 − (sec A – tan A)) (sec A – tan A) (1 + (sec A + tan A))
= (sec2A − tan2A) (1 – sec A + tan A) (1 + sec A + tan A)
=2 tan A
(1 + cot A - cosec A)(1 + tan A + sec A) = 2
LHS = (1 + cot A - cosec A) (1 + tan A + sec A)
(cosec θ – sec θ) (cot θ – tan θ) = (cosec θ + sec θ) (sec θ cosec θ − 2)
LHS = (cosec θ – sec θ) (cot θ – tan θ)
RHS = (cosec θ + sec θ) (sec θ cosec θ - 2)
= cosec A - sec A
= RHS = LHS Hence proved.
= LHS = RHS Hence proved.
= sin A × cos3 A +cos A × sin3A
= sin A cos A (cos2A + sin2A)
= sin A cos A
sec4A (1 − sin4A) – 2 tan2A = 1
Given, L.H.S = sec4A (1 - sin4A) - 2tan2A
= sec4A − sec4A × sin4A − 2tan4A
= sec4A - tan4A – 2 tan4A
= (sec2A)2 - tan4A – 2 tan4A
= (1+ tan2A)2 − tan4A − 2tan4A
= 1 + tan4A + 2tan2A − tan4A − 2tan4A
Here, sin2A + cos2A = 1
Solving, RHS ⟹
Multiplying Nr. And Dr. with (1 + Sin A)
Hence, LHS = RHS
= (1 + cot A + tan A) (sin A – cos A)
⟹ sin A – cos A + cot A sin A – cot A cos A + sin A tan A – tan A cos A
⟹ sin A tan A – cos A cot A
If cosec θ – sin θ = a3, sec θ – cos θ = b3, Prove that a2b2 (a2+ b2) = 1
Given, cosec θ – sin θ = a3
Here cos2θ A + sin2 A = 1
Squaring on both sides
If a cos3 θ + 3 a cos θ sin2θ = m, a sin3θ + 3 a cos2θ sin θ = n,
Substitute the values of m and n in the above equation
⟹ a cos3θ + 3 a cos θ sin2 θ + a sin3 θ + 3 a cos2θ sin2/3 θ + a cos3 θ + 3 a cos θ sin2 θ – a sin3 θ – 3 a cos2 θ sin2/3 θ
⟹ (a)2/3 cos3θ + 3 cos θ sin2 θ + sin3θ + 3 cos2 θ sin2/3θ + (a)2/3 cos3 θ + 3 cos θ sin 2θ - sin3 θ – 3 cos2 θ sin2/3θ
⟹ (a)2/3((cos θ + sin θ)3)2/3 + (a)2/3 ((cos θ – sin θ)3)2/3
⟹ (a)2/3[(cos θ + sin θ)2] + (a)2/3[(cos θ – sin θ)2]
⟹ (a)2/3((cos2θ + sin2θ + 2 sin θ cos θ)) + (a)2/3((cos2θ + sin2θ – 2 sin θ cos θ))
⟹ (a)2/3[ 1 + 2 sin θ cos θ] + (a)2/3[1 – 2 sin θ cos θ]
⟹ (a)2/3[1 + 2 sin θ cos θ] + 1 – 2 sin θ cos θ]
⟹ (a)2/3(1 + 1)
⟹ 2(a)2/3
If x = a cos3θ, y = b sin3θ,
x = a cos3 θ: y = b sin3 θ
If a cos θ + b sin θ = m and a sin θ – b cos θ = n, Prove that a2 + b2 = m2 + n2
R.H.S = m2 + n2 = m2 + n2
= (a cos θ + b sin θ)2 + (a sin θ – b cos θ)2
= a2 cos2 θ + b2 sin2θ + 2ab sin θ cos θ + a2 + sin2θ + b2 cos2θ – 2ab sin θ cos θ
= a2 cos2θ + b2 cos2θ + b2 sin2θ + a2 sin2θ
= a2(sin2 θ + cos2 θ) + b2(sin2 θ + cos2 θ)
= a2 + b2 (∵ sin2 θ + cos2θ = 1)
If cos A + cos2A = 1, Prove that sin2A + sin4A = 1
Given - cos A + cos2 A = 1
We have to prove sin2 A + sin4 A = 1
Now, cos A + cos2 A = 1
cos A = 1 - cos2 A
cos A = sin2 A
sin2 A = cos A
We have sin2 A + sin4 A = cos A + (cos A)2
= cos A + cos2 A = 1
If cos θ + cos2 θ = 1, Prove that sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2 = 1
cos θ + cos2 θ = 1
cos θ = 1 − cos2 θ
cos θ = sin2 θ (i)
Now, sin12 θ + 3 sin10 θ + 3 sin8 θ + sin6 θ + 2 sin4 θ + 2 sin2 θ − 2
= (sin4 θ)3 + 3 sin4 θ. sin2 θ (sin4 θ + sin2 θ) + (sin2 θ)3 + 2(sin2 θ)2 + 2 sin2 θ − 2
Using (a + b)3 = a3 + b3 + 3 ab(a + b) and also from(i) cos θ = sin2 θ
(sin4 θ + sin2 θ)3 + 2cos2 θ + 2 cos θ − 2
((sin2 θ)2 + sin2 θ)3 + 2 cos2 θ + 2 cos θ – 2
(cos2 θ + sin2 θ)3 + 2 cos2 θ + 2 cos θ − 2
1 + 2(cos2 θ + sin2 θ) − 2
= 1 + 2(1) −2
L.H.S = R.H.S Hence proved.
Given that: (1 + cos α)(1 + cos β)(1 + cos γ) = (1 – cos α)(1 – cos β)(1 – cos γ). Show that one of the values of each member of this equality is sin α sin β sin γ.
Multiply (i) with sin α sin β sin γ and divide it with same we get
Hence sin α sin β sin γ is the member of equality.
If sin θ + cos θ = x,
sin θ + cos θ = x
(sin θ + cos θ)2 = x2
⇒ sin2θ + cos2θ + 2 sin θ cos θ = x2
We know that sin2 θ + cos2 θ = 1
Cubing on both sides (sin2 θ + cos2 θ)3 = 13
sin6 θ + cos6 θ + 3 sin2 θ cos2 θ (sin2 θ + cos2 θ) = 1
⇒ sin6 θ + cos6 θ = 1 – 3 sin2 θ cos2θ
If x = a sec θ cos ϕ, y = b sec θ sin ϕ and z = c tan ϕ,
Given, x = a sec θ cos ϕ, y = b sec θ sin ϕ, z = c tan ϕ squaring x, y, z on the sides
x2 = a2sec2θ cos2 ϕ x2/a2 = sec2θ cos2 ϕ --- 1
y2 = b2 sec2θ sin2 ϕ y2/b2 = sec2θ sin2 ϕ --- 2
z2 = c2 tan2 ϕ z2/c2 = tan2 ϕ --- 3
Substitute equation 1, 2, 3 in x2/a2 + y2/b2 − z2/c2
⟹ x2/a2 + y2/b2 − z2/c2
⟹ Sec2 θ cos2 ϕ + sec2 θ sin2 ϕ - tan2 ϕ
⟹ sec2θ(cos2 ϕ + sin2 ϕ) - tan2 ϕ
cos2 ϕ + sin2 ϕ = 1
⟹ sec2θ (1) - tan2 ϕ and sec2θ − tan2θ = 1
Hence, L.H.S= R.H.S
If sin θ + 2 cos θ. Prove that 2 sin θ – cos θ = 2
Given, sin θ + 2 cos θ = 1
⟹ (sin θ + 2 cos θ)2 = 12
⟹ sin2θ + 4 cos2θ + 4 sin θ cos θ = 1
⟹ 4 cos2 θ + 4 sin θ cos θ = 1 - sin2θ
Here, 1 - sin2θ = cos2θ
⟹ 4 cos2θ + 4 sin θ cos θ - cos2θ = 0
⟹ 3 cos2θ + 4 sin θ cos θ = 0 ---- 1
We have,
2 sin θ – cos θ = 2
Squaring L.H.S
(2 sin θ – cos θ)2
= 4 sin2θ + cos2θ – 4 sin θ cos θ
Here, 4 sin θ cos θ = 3 cos2θ
= 4 sin2θ + cos2θ + 3 cos2θ
= 4 sin2 θ + 4 cos2 θ
= 4(sin2θ + cos2θ)
= 4(1)
= 4 (2 sin θ – cos θ)2 = 4
⟹ 2sin θ – cos θ
= 2
Signing up with Facebook allows you to connect with friends and classmates already using askIItians. It’s an easier way as well. “Relax, we won’t flood your facebook news feed!”
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
Chapter 6: Trigonometric Identities Exercise...