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USE CODE: SELF10

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: 69



Solution:



= cosec A - sec A 

= RHS = LHS Hence proved.

Question: 70



Solution:



= LHS = RHS Hence proved.

Question: 71



Solution:



= sin A × cos3 A +cos A × sin3A 

= sin A cos A (cos2A + sin2A) 

= sin A cos A

LHS = RHS Hence proved.

Question: 72

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: 73



Solution:



Here, sin2A + cos2A = 1



Solving, RHS ⟹



Multiplying Nr. And Dr. with (1 + Sin A)



Hence, LHS = RHS

Question: 74



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: 75



Solution:



Question: 76

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: 77

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: 78

If x = a cos3θ, y = b sin3θ,



Solution:

x = a cos3 θ: y = b sin3 θ 



= 1

Hence proved.

Question: 79

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: 80

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: 81

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: 82

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: 83

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: 84

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: 85

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

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