Chapter 7 Trigonometric Identities Ex 7.1

RBSE Solutions for Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.1.

Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.1 Solution is provided in this post. Here we have provide the solutions of RBSE Boards Books according to chapter wise.

Chapter 7 Trigonometric Identities Ex 7.1

RBSE Solutions for Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.1.

Question 1.
For ∠θ express all the Trigonometric ratios in terms of sec θ.
Solution :
(i) sin2 θ + cos2 θ = 1
⇒ sin2 θ = 1 – cos2 θ

(ii) Cos θ = 
(iii) ∵ 1 + tan2 θ = sec2 θ
or, tan2 θ = sec2 θ – 1
or, tan θ = 
(iv) cot θ = 
= 
(v) cosec θ = 

Chapter 7 Trigonometric Identities Ex 7.1

Question 2.
Express trigonometric sin θ, sec θ, and tan θ in terms of cot θ.
Solution :
(i) sin θ = 

(ii) sec2 θ = 1 + tan2 θ

Question 3.
Verify the following with the help of identities.
cos2θ + cos2θ cot2θ = cot2θ
Solution :
L.H.S. = cos2θ + cos2θ cot2θ
= cos2θ(1 + cot2θ)

= cot2θ
= R.H.S.

Question 4.
sec θ (1 – sin θ) (sec θ + tan θ) = 1
Solution :
L.H.S. = sec θ(1 – sinθ) (secθ + tanθ)

= 1
= R.H.S

Question 5.
cosec2 θ + sec2 θ = cosec2 θ sec2 θ
Solution :
LH.S. = cosec2 θ + sec2 θ

= cosec2 θ sec2 θ
= R.H.S

Question 6.
 = sec θ – tan θ
Solution :
L.H.S = 
Multiply numerator and denominator by (1 – sin θ)

= 
= sec θ – tan θ
= R.H.S.

Question 7.
 = tan θ + cot θ
Solution :
L.H.S. = 

= tan θ + cot θ
= R.H.S.

Question 8.
 = tan α tan β
Solution :
L.H.S. = 

= tan α tan β
= R.H.S.

Question 9.
 = 2 sec θ
Solution :
L.H.S = 

= 2 sec θ
= R.H.S.

Chapter 7 Trigonometric Identities Ex 7.1

Question 10.
 = 1
Solution :
L.H.S. = 

= 1
= R.H.S.

Question 11.
cot θ – tan θ = 
Solution :
L.H.S. = cot θ – tan θ

= 
= R.H.S.

Question 12.
cos4 θ + sin4 θ = 1 – 2cos2 θ sin2 θ
Solution :
L.H.S. = cos4 θ + sin4 θ
= (cos2 θ)2 + (sin2 θ)2 + 2sin2 θ. cos2 θ – 2 sin2 θ cos2 θ
= (sin2 θ + cos2 θ)2 – 2sin2 θ cos2 θ
= 1 – 2sin2 θ cos2 θ [∵ sin2 θ + cos2 θ = 1]
= R.H.S

Question 13.
(sec θ – cos θ) (cot θ + tan θ) = tan θ sec θ
solution :
L.H.S. = (sec θ – cos θ) (cot θ + tan θ)

= tan θ sec θ
= R.H.S.

Question 14.
 = tan2 α
L.H.S. = 

= R.H.S.

Question 15.
 = 
Solution :
L.H.S. = 
Multiply numerator and denominator by (1 + cos θ)
= 

= 
= R.H.S.

Chapter 7 Trigonometric Identities Ex 7.1

Question 16.
sin6 θ + cos6 θ = 1 – 3sin2 θ cos2 θ
Solution :
L.H.S. = (sin2 θ)3 + (cos2 θ)2
= (sin2 θ + cos2 θ)
(sin2 θ + cos2 θ – sin2 θ cos2 θ) [∵ a3 + b3 = (a + b) (a2 + b2 – ab)]
= sin4 θ + cos4 θ – sin2 θ cos2 θ
= (sin2 θ)2 + (cos2 θ)2 + 2sin2 θ cos2 θ
= 2 sin2 θ cos2 θ – sin2 θ cos2 θ
= (sin2 θ + cos2 θ)2 – 3sin2 θ cos2 θ
= 1 – 3sin2 θ cos2 θ
= R.H.S.

Question 17.
 = 1 + tan θ + cot θ
Solution :
L.H.S. = 


= 1 + tan θ + cot θ
= R.H.S.
Alternate :
L.H.S. = 

= tan θ + 1 + cot θ
= 1 + tan θ + cot θ
= R.H.S.

Question 18.
sin θ(1 + tan θ) + cos θ(1 + cot θ) = cosec θ + sec θ
Solution :
L.H.S. = sin θ(1 + tan θ) + cos θ(1 + cot θ)


= cosec θ + sec θ
= R.H.S.

Question 19.
sin2 θ cos θ + tan θ sin θ + cos3 θ = sec θ.
Solution :
L.H.S. = sin2 θ cos θ + tan θ sin θ + cos3 θ
= (sin2 θ cos θ + cos3 θ) + tan θ sin θ
= cos θ(sin2 θ + cos2 θ) + tan θ sin θ

= sec θ
= R.H.S.

Question 20.
 = 1 + sec θ cosec θ
Solution :
L.H.S. = 


= 1 + sec θ cosec θ
= R.H.S.

Chapter 7 Trigonometric Identities Ex 7.1

Question 21.
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A
Solution :
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2


= 5 + sec2 A + cosec2 A
= 5 + (1 + tan2 A) + (1 + cot2 A)
= 7 + tan2 A + cot2 A
R.H.S.

Question 22.
sin8θ – cos8θ = (sin2 θ – cos2 θ) (1 – 2sin2 θ cos2 θ)
Solution :
L.H.S. = sin8 θ – cos2 θ
(sin4 θ)2 – (cos4 θ)2
= (sin4 θ + cos4 θ) (sin4 θ – cos4 θ)
= [(sin2 θ)2 + (cos2 θ)2 + 2cos2 θ sin2 θ – 2sin2 θ cos2 θ] [(sin2 θ)2 – (cos2 θ)2]
= [(sin2 θ + cos2 θ) – 2sin2 θ cos2 θ] [(sin2 θ + cos2 θ) (sin2 θ – cos2 θ)]
= (1 – 2Sin2 θ cos2 θ) (sin2 θ – cos2 θ)
= (sin2 θ – cos2 θ) (1 – 2sin2 θ cos2 θ)
= R.H.S.

Question 23.
 = cot θ + cosec θ
Solution :
L.H.S. = 

Multiply numerator and denominator by (1 + cos θ)

= cosec θ + cot θ or cot θ + cosec θ
= R.H.S.

Question 24.

= sin2 θ cos2 θ
Solution :
L.H.S.

= sin2 θ cos2 θ
= R.H.S.

Chapter 7 Trigonometric Identities Ex 7.1

Question 25.

Solution :
L.H.S.

= R.H.S.
Again by equation (i)

= R.H.S.

Question 26.
 = sin A + cos A.
Solution :
L.H.S. = 


= sin A + cos A
= R.H.S

Question 27.
(cosec A – sin A) (sec A – cos A) = 
Solution :
L.H.S. = (cosec A – sin A) (sec A – cos A)

= 
= R.H.S.

Question 28.

= 1 + sin θ cos θ
Solution :
L.H.S.

= 1 + sin θ cos θ [∵ sin2 + cos2 = 1]
= R.H.S.

Question 29.
If sec θ + tan θ = p then prove that  = sin θ
Solution :
Given, sec θ + tan θ = p
L.H.S. = 


= sin θ
= R.H.S.

Question 30.
If  = m and  = n then prove that (m2 + n2) cos2 B = n2
Solution :
Given m =  and n = 
L.H.S. = (m2 + n2) cos2B

= n2
= R.H.S.

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