**RBSE Solutions for Class 10 Maths Chapter 7 Trigonometric Identities** Ex 7.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 7 Trigonometric Identities Exercise 7.2.

## Rajasthan Board RBSE Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.2

Question 1.

Find the value of (RBSESolutions.com) the following :

Solution :

Question 2 :

Find the value (RBSESolutions.com) of the following :

(i) cosec 25° – sec 25°

(ii) cot 34° – tan 56°

(iii) \(\frac { { sin36 }^{ \circ } }{ { cos54 }^{ \circ } } -\frac { { sin54 }^{ \circ } }{ { cos36 }^{ \circ } } \)

(iv) sin θ cos(90° – θ) + cos θ sin (90° – θ)

Solution :

(i) cosec 25° – sec 65°

= cosec(90° – 65°) – sec 65° [∵ cosec(90° – θ) = sec θ]

sec 65° – sec 65°

= 0

(ii) cot 34° – tan 56°

= cot(90° – 56°) – tan 56° [∵ cot(90 – θ) = tan θ]

= tan 56° – tan 56°

= 0

(iii) \(\frac { { sin36 }^{ \circ } }{ { cos54 }^{ \circ } } -\frac { { sin54 }^{ \circ } }{ { cos36 }^{ \circ } } \)

= 1 – 1

= 0

(iv) sin θ cos(90° – θ) + cos θ sin (90° – θ)

= sin θ sin θ + cos θ cos θ [∵ cos (90° – θ) = sin θ sin (90° – θ) = cos θ]

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

= 1

Question 3.

(i) sin 70° sin 20° – cos 20° cos 70°

(ii) \(\frac { { 2cos67 }^{ \circ } }{ { sin23 }^{ \circ } } -\frac { { tan40 }^{ \circ } }{ { cot50 }^{ \circ } } \) – cos 60°

Solution :

(i) sin 70° sin 20° – cos 20° cos 70°

= sin(90° – 20°)sin 20° – cos 20° cos(90° – 20°)

= cos 20° sin 20° – cos 20° sin 20°

= 0

(ii) \(\frac { { 2cos67 }^{ \circ } }{ { sin23 }^{ \circ } } -\frac { { tan40 }^{ \circ } }{ { cot50 }^{ \circ } } \) – cos 60°

= 2 – 3/2

= \(\frac { 1 }{ 2 }\)

Question 4.

Solution :

Question 5.

(i) tan 12° cot 38° cot 52° cot 60° tan 78°

(ii) tan 5° tan 25° tan 30° tan 45° tan 65° tan 85°

Solution :

(i) tan 12° cot 38° cot 52° cot 60° tan 78°

= tan 12° cot 38° cot(90° – 38°) cot 60° tan(90° – 12°)

= tan 12° cot 38° tan 38° cot 60° cot 12° [∵ cot(90° – θ) = tan θ]

= cot 60° = \(\frac { 1 }{ \sqrt { 3 } }\)

(ii) tan 5° tan 25° tan 30° tan 45° tan 65° tan 85°

= tan 5° tan 25° tan 30° × 1 × tan(90° – 25°) tan(90° – 5°)

= tan 5° tan 25° tan 30° × cot 25° cot 5° [∵ tan(90° – θ) = cot θ]

Question 6.

Express the following (RBSESolutions.com) terms of trigonometric ratios of angles 0° to 45°.

(i) sin 81° + sin 71°

(ii) tan 68° + sec 68°

Solution :

(i) sin 81° + sin 71°

= sin(90° – 9°) + sin(90° – 19°)

= cos 9° + cos 19°

(ii) tan 68° + sec 68°

= tan(90° – 22°) + sec(90° – 22°)

= cot 22° + cosec 22°

Question 7.

Verify the following –

sin 65° + cos 25° = 2 cos 25°

Solution :

L.H.S. = sin 65° + cos 25°

= sin(90° – 25°) + cos 25°

cos 25° + cos 25° [∵ sin(90 – θ) = cos θ]

= 2 cos 25°

= R.H.S.

Question 8.

sin 35° sin 55° – cos 35° cos 55° = 0

Solution :

L.H.S. = sin 35° sin 55° – cos 35° cos 55°

= sin 35° sin(90° – 35°) – cos 35 cos(90° – 35°)

= sin 35° cos 35° – cos 35° sin 35°

∵ sin(90° – θ) = cos θ

cos(90° – θ) = sin θ

= 0

= R.H.S.

Question 9.

\(\frac { { cos70 }^{ \circ } }{ { sin20 }^{ \circ } } +\frac { { cos59 }^{ \circ } }{ { sin31 }^{ \circ } } \) – 8sin^{2}30° = 0

Solution :

L.H.S. = \(\frac { { cos70 }^{ \circ } }{ { sin20 }^{ \circ } } +\frac { { cos59 }^{ \circ } }{ { sin31 }^{ \circ } } \) – 8sin^{2}30°

2 – 2 = 0

= R.H.S.

Question 10.

sin (90° – θ) cos(90° – θ) = \(\frac { tan\theta }{ { 1+tan }^{ 2 }\theta }\)

Solution :

L.H.S. = sin (90° – θ) cos(90° – θ)

= cos θ sin θ

= R.H.S.

Question 11.

Solution :

L.H.S.

= cos^{2} θ + sin^{2} θ

= 1 [∵ cos^{2} θ + sin^{2} θ = 1]

= R.H.S.

Question 12.

Solution :

L.H.S.

= cos^{2} θ – cos^{2} θ

= 0

= R.H.S.

Question 13.

Solution :

L.H.S.

= sin^{2} θ

= R.H.S.

Question 14.

Solution :

L.H.S.

= sin^{3} θ cos θ + cos^{3} θ sin θ

= sin θ cos θ (sin^{2} θ + cos^{2} θ) [∵ sin^{2} θ + cos^{2} θ = 1]

= sin θ cos θ

= R.H.S.

Question 15.

If sin 3θ = cos(θ – 6°) here (RBSESolutions.com) 3θ and (θ – 6°) are acute angles, then find the value of θ.

Solution :

Given :

sin 3θ = cos(θ – 6°)

or cos(90° – 3θ) = cos(θ – 6°) [∵ cos(90° – θ) = sin θ]

or 90° – 3θ = θ – 6°

or 3θ + θ = 90° + 6°

or 4θ = 96°

or θ = \(\frac { 96 }{ 4 }\) = 24°

Question 16.

If sec 5θ = cosec(θ – 36°) here 5θ is an (RBSESolutions.com) acute angle, then find the value of θ.

Solution :

Given :

sec 5θ = cosec(θ – 36°)

or cosec(90° – 5θ) = cosec(θ – 36°) [∵ cosec(90° – θ) = sec θ]

or 900 – 5θ = θ – 36°

or 5θ + θ = 90° + 36°

or 6θ = 126°

or θ = \(\frac { 126 }{ 6 }\) = 21°

Thus θ = 21°

Question 17.

If A, B and C are interior (RBSESolutions.com) angles of a triangle ABC then Prove that

Solution :

we know that, in any triangle

∠A + ∠B + ∠C = 180°

Question 18.

If cos 2θ = sin 4θ and 2θ and 4θ are acute (RBSESolutions.com) angles then find θ.

Solution :

Given :

cos 2θ = sin 4θ

or cos 2θ = cos(90° – 4θ) [∵ cos(90° – θ) = sin θ]

or 2θ = 90° – 4θ

or 6θ = 90°

or θ = \(\frac { { 90 }^{ \circ } }{ 6 }\)

θ = 15°

We hope the given RBSE Solutions for Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 7 Trigonometric Identities Exercise 7.2, drop a comment below and we will get back to you at the earliest.

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