RBSE Solutions for Class 10 Maths Chapter 7 Trigonometric Identities Ex 7.2

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]
= sin² θ + cos² θ [∵ sin² θ + cos² 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.
\(\frac { { cos70 }^{ \circ } }{ { sin20 }^{ \circ } } +\frac { { cos59 }^{ \circ } }{ { sin31 }^{ \circ } } \) – 8sin²30° = 0
Solution :
L.H.S. = \(\frac { { cos70 }^{ \circ } }{ { sin20 }^{ \circ } } +\frac { { cos59 }^{ \circ } }{ { sin31 }^{ \circ } } \) – 8sin²30°

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.

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

Question 12.

Solution :
L.H.S.


= cos² θ – cos² θ
= 0
= R.H.S.

Question 13.

Solution :
L.H.S.

= sin² θ
= R.H.S.

Question 14.

Solution :
L.H.S.

= sin³ θ cos θ + cos³ θ sin θ
= sin θ cos θ (sin² θ + cos² θ) [∵ sin² θ + cos² 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.