**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) sin^{2} θ + cos^{2} θ = 1

⇒ sin^{2} θ = 1 – cos^{2} θ

(ii) Cos θ =

(iii) ∵ 1 + tan^{2} θ = sec^{2} θ

or, tan^{2} θ = sec^{2} θ – 1

or, tan θ =

(iv) cot θ =

=

(v) cosec θ =

Question 2.

Express (RBSESolutions.com) trigonometric sin θ, sec θ, and tan θ in terms of cot θ.

Solution :

(i) sin θ =

(ii) sec^{2} θ = 1 + tan^{2} θ

(iii) tan θ =

Question 3.

Verify the following (RBSESolutions.com) with the help of identities.

cos^{2}θ + cos^{2}θ cot^{2}θ = cot^{2}θ

Solution :

L.H.S. = cos^{2}θ + cos^{2}θ cot^{2}θ

= cos^{2}θ(1 + cot^{2}θ)

= cot^{2}θ

= 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.

cosec^{2} θ + sec^{2} θ = cosec^{2} θ sec^{2} θ

Solution :

LH.S. = cosec^{2} θ + sec^{2} θ

= cosec^{2} θ sec^{2} θ

= R.H.S

Question 6.

= sec θ – tan θ

Solution :

L.H.S =

Multiply numerator (RBSESolutions.com) 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.

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.

cos^{4} θ + sin^{4} θ = 1 – 2cos^{2} θ sin^{2} θ

Solution :

L.H.S. = cos^{4} θ + sin^{4} θ

= (cos^{2} θ)^{2} + (sin^{2} θ)^{2} + 2sin^{2} θ. cos^{2} θ – 2 sin^{2} θ cos^{2} θ

= (sin^{2} θ + cos^{2} θ)^{2} – 2sin^{2} θ cos^{2} θ

= 1 – 2sin^{2} θ cos^{2} θ [∵ sin^{2} θ + cos^{2} 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.

= tan^{2} α

L.H.S. =

= R.H.S.

Question 15.

=

Solution :

L.H.S. =

Multiply numerator (RBSESolutions.com) and denominator by (1 + cos θ)

=

=

= R.H.S.

Question 16.

sin^{6} θ + cos^{6} θ = 1 – 3sin^{2} θ cos^{2} θ

Solution :

L.H.S. = (sin^{2} θ)^{3} + (cos^{2} θ)^{2}

= (sin^{2} θ + cos^{2} θ)

(sin^{2} θ + cos^{2} θ – sin^{2} θ cos^{2} θ) [∵ a^{3} + b^{3} = (a + b) (a^{2} + b^{2} – ab)]

= sin^{4} θ + cos^{4} θ – sin^{2} θ cos^{2} θ

= (sin^{2} θ)^{2} + (cos^{2} θ)^{2} + 2sin^{2} θ cos^{2} θ

= 2 sin^{2} θ cos^{2} θ – sin^{2} θ cos^{2} θ

= (sin^{2} θ + cos^{2} θ)^{2} – 3sin^{2} θ cos^{2} θ

= 1 – 3sin^{2} θ cos^{2} 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.

sin^{2} θ cos θ + tan θ sin θ + cos^{3} θ = sec θ.

Solution :

L.H.S. = sin^{2} θ cos θ + tan θ sin θ + cos^{3} θ

= (sin^{2} θ cos θ + cos^{3} θ) + tan θ sin θ

= cos θ(sin^{2} θ + cos^{2} 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.

Question 21.

(sin A + cosec A)^{2} + (cos A + sec A)^{2} = 7 + tan^{2}A + cot^{2}A

Solution :

L.H.S. = (sin A + cosec A)^{2} + (cos A + sec A)^{2}

= 5 + sec^{2} A + cosec^{2} A

= 5 + (1 + tan^{2} A) + (1 + cot^{2} A)

= 7 + tan^{2} A + cot^{2} A

R.H.S.

Question 22.

sin^{8}θ – cos^{8}θ = (sin^{2} θ – cos^{2} θ) (1 – 2sin^{2} θ cos^{2} θ)

Solution :

L.H.S. = sin^{8} θ – cos^{2} θ

(sin^{4} θ)^{2} – (cos^{4} θ)^{2}

= (sin^{4} θ + cos^{4} θ) (sin^{4} θ – cos^{4} θ)

= [(sin^{2} θ)^{2} + (cos^{2} θ)^{2} + 2cos^{2} θ sin^{2} θ – 2sin^{2} θ cos^{2} θ] [(sin^{2} θ)^{2} – (cos^{2} θ)^{2}]

= [(sin^{2} θ + cos^{2} θ) – 2sin^{2} θ cos^{2} θ] [(sin^{2} θ + cos^{2} θ) (sin^{2} θ – cos^{2} θ)]

= (1 – 2Sin^{2} θ cos^{2} θ) (sin^{2} θ – cos^{2} θ)

= (sin^{2} θ – cos^{2} θ) (1 – 2sin^{2} θ cos^{2} θ)

= R.H.S.

Question 23.

= cot θ + cosec θ

Solution :

L.H.S. =

Multiply numerator (RBSESolutions.com) and denominator by (1 + cos θ)

= cosec θ + cot θ or cot θ + cosec θ

= R.H.S.

Question 24.

= sin^{2} θ cos^{2} θ

Solution :

L.H.S.

= sin^{2} θ cos^{2} θ

= R.H.S.

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 [∵ sin^{2} + cos^{2} = 1]

= R.H.S.

Question 29.

If sec θ + tan θ = p then (RBSESolutions.com) 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 (m^{2} + n^{2}) cos^{2} B = n^{2}

Solution :

Given m = and n =

L.H.S. = (m^{2} + n^{2}) cos^{2}B

= n^{2}

= 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.

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