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

RBSE Solutions for Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.1 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 7 Trigonometric Identities Exercise 7.1.

Rajasthan Board RBSE Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.1

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

(ii) Cos θ = \(\frac { 1 }{ sec\theta }\)
(iii) ∵ 1 + tan² θ = sec² θ
or, tan² θ = sec² θ – 1
or, tan θ = \(\sqrt { { sec }^{ 2 }\theta -1 }\)
(iv) cot θ = \(\frac { 1 }{ tan\theta }\)
= \(\frac { 1 }{ \sqrt { { sec }^{ 2 }\theta -1 } }\)
(v) cosec θ = \(\frac { 1 }{ sin\theta }\)

Question 2.
Express (RBSESolutions.com) trigonometric sin θ, sec θ, and tan θ in terms of cot θ.
Solution :
(i) sin θ = \(\frac { 1 }{ cosec\theta }\)

(ii) sec² θ = 1 + tan² θ

(iii) tan θ = \(\frac { 1 }{ cot\theta }\)

Question 3.
Verify the following (RBSESolutions.com) 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.
\(\sqrt { \frac { 1-sin\theta }{ 1+sin\theta } }\) = sec θ – tan θ
Solution :
L.H.S = \(\sqrt { \frac { 1-sin\theta }{ 1+sin\theta } }\)
Multiply numerator (RBSESolutions.com) and denominator by (1 – sin θ)

= \(\frac { 1 }{ cos\theta } -\frac { sin\theta }{ cos\theta }\)
= sec θ – tan θ
= R.H.S.

Question 7.
\(\sqrt { { sec }^{ 2 }\theta +{ cosec }^{ 2 }\theta } \) = tan θ + cot θ
Solution :
L.H.S. = \(\sqrt { { sec }^{ 2 }\theta +{ cosec }^{ 2 }\theta } \)

= tan θ + cot θ
= R.H.S.

Question 8.
\(\frac { tan\alpha +tan\beta }{ cot\alpha +cot\beta }\) = tan α tan β
Solution :
L.H.S. = \(\frac { tan\alpha +tan\beta }{ cot\alpha +cot\beta }\)

= tan α tan β
= R.H.S.

Question 9.
\(\frac { 1+sin\theta }{ cos\theta } +\frac { cos\theta }{ 1+sin\theta }\) = 2 sec θ
Solution :
L.H.S = \(\frac { 1+sin\theta }{ cos\theta } +\frac { cos\theta }{ 1+sin\theta }\)

= 2 sec θ
= R.H.S.

Question 10.
\(\frac { { sin }^{ 4 }\theta -{ cos }^{ 4 }\theta }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }\) = 1
Solution :
L.H.S. = \(\frac { { sin }^{ 4 }\theta -{ cos }^{ 4 }\theta }{ { sin }^{ 2 }\theta -{ cos }^{ 2 }\theta }\)

= 1
= R.H.S.

Question 11.
cot θ – tan θ = \(\frac { { 1-2sin }^{ 2 }\theta }{ sin\theta cos\theta }\)
Solution :
L.H.S. = cot θ – tan θ

= \(\frac { { 1-2sin }^{ 2 }\theta }{ sin\theta cos\theta }\)
= 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.
\(\frac { { 1-tan }^{ 2 }\alpha }{ { cot }^{ 2 }\alpha -1 }\) = tan² α
L.H.S. = \(\frac { { 1-tan }^{ 2 }\alpha }{ { cot }^{ 2 }\alpha -1 }\)

= R.H.S.

Question 15.
\(\frac { sin\theta }{ 1-cos\theta }\) = \(\frac { 1+cos\theta }{ sin\theta }\)
Solution :
L.H.S. = \(\frac { sin\theta }{ 1-cos\theta }\)
Multiply numerator (RBSESolutions.com) and denominator by (1 + cos θ)
= \(\frac { sin\theta }{ 1-cos\theta } \times \frac { 1+cos\theta }{ 1+cos\theta }\)

= \(\frac { 1+cos\theta }{ sin\theta }\)
= R.H.S.

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.
\(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \) = 1 + tan θ + cot θ
Solution :
L.H.S. = \(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \)


= 1 + tan θ + cot θ
= R.H.S.
Alternate :
L.H.S. = \(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \)

= 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.
\(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \) = 1 + sec θ cosec θ
Solution :
L.H.S. = \(\frac { tan\theta }{ 1-cot\theta } +\frac { cot\theta }{ 1-tan\theta } \)


= 1 + sec θ cosec θ
= R.H.S.

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.
\(\sqrt { \frac { sec\theta +1 }{ sec\theta -1 } }\) = cot θ + cosec θ
Solution :
L.H.S. = \(\sqrt { \frac { sec\theta +1 }{ sec\theta -1 } }\)

Multiply numerator (RBSESolutions.com) 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.

Question 25.

Solution :
L.H.S.

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

= R.H.S.

Question 26.
\(\frac { cosA }{ 1-tanA } +\frac { SinA }{ 1-cotA }\) = sin A + cos A.
Solution :
L.H.S. = \(\frac { cosA }{ 1-tanA } +\frac { SinA }{ 1-cotA }\)


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

Question 27.
(cosec A – sin A) (sec A – cos A) = \(\frac { 1 }{ tanA+cotA }\)
Solution :
L.H.S. = (cosec A – sin A) (sec A – cos A)

= \(\frac { 1 }{ tanA+cotA }\)
= 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 (RBSESolutions.com) prove that \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }\) = sin θ
Solution :
Given, sec θ + tan θ = p
L.H.S. = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 }\)


= sin θ
= R.H.S.

Question 30.
If \(\frac { cosA }{ cosB }\) = m and \(\frac { cosA }{ sinB }\) = n then prove that (m² + n²) cos² B = n²
Solution :
Given m = \(\frac { cosA }{ cosB }\) and n = \(\frac { cosA }{ sinB }\)
L.H.S. = (m² + n²) cos²B

= n²
= R.H.S.

We hope the given RBSE Solutions for Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.1 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 7 Trigonometric Identities Exercise 7.1, drop a comment below and we will get back to you at the earliest.