Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 8 Introduction to Trigonometry Exercise 8.3

Question 1.

Evaluate:

(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)

(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)

(iii) cos 48° – sin 42°

(iv) cosec 31° – sec 59°.

Solution:

[∵ cos (90° – θ) = sin θ] tan 26°

(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\) = \(\frac{\tan 26^{\circ}}{\cot \left(90^{\circ}-26^{\circ}\right)}\)

= \(\frac{\tan 26^{\circ}}{\tan 26^{\circ}}\) = 1

[∵ cot (90° – θ) = tan θ]

(iii) cos 48° – sin 42°

= cos (90° – 42°) – sin 42°

[∵ cos (90° – θ) = sin θ]

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

= cosec 31° – sec (90° – 31°)

= cosec 31° – cosec 31°

[∵ sec (90° – θ) = cosec θ] = 0

Question 2.

Show that :

(i) tan 48° tan 23° tan 42° tan 67° = 1

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Solution:

(i) L.H.S. = tan 48° tan 23° tan42° tan 67°

= tan 48° × tan 23° × tan (90° – 48°) × tan (90° – 23°)

= tan 48° × tan 23° × cot 48° × cot 23°

∵ tan (90° – θ) = cot θ

= tan 48° × tan23° × \(\frac{1}{\tan 48^{\circ}}\) × \(\frac{1}{\tan 23^{\circ}}\) = 1

[∵ cot θ = \(\frac{1}{\tan θ^{\circ}}\) ]

∴ L.H.S. = R.H.S.

(ii) L.H.S.= cos 38° cos 52° – sin 38° sin 52°

= cos 38° × cos (90° – 38°) – sin 38° × sin (90° – 38°)

= cos 38° × sin 38° – sin 38° × cos 38°

[∵ cos (90° – θ) = sin θ

sin (90° – θ) = cos θ]

= 0

∴ L.H.S. = R.H.S.

Question 3.

If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

According to the question tan 2A = cot (A – 18°)

To find A we must have either cot θ or tan θ on both sides.

[∵ cot (90° – θ) = tan θ]

⇒ cot (90° – 2A) = cot (A – 18°)

⇒ 90° – 2A = A – 18°

⇒ 3A = 108°

⇒ A = 36°

Question 4.

If tan A = cot B, prove that A + B = 90°

Solution:

According to the question

tan A = cot B

To show that A + B = 90°, we must have either tan 0 or cot 0 on both sides.

[∵ tan (90° – θ) = cot θ]

⇒ tan A = tan (90° – B)

⇒ A = 90° – B

⇒ A + B = 90°

Question 5.

If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Solution:

According to the question

sec 4A = cosec (A – 20°)

To determine A we must have either sec θ or cosec θ on both sides.

[∵ cosec (90° – θ) = sec θ]

⇒ cosec (90° – 4A) = cosec (A – 20°)

⇒ 90° – 4A = A – 20°

⇒ 5A = 110°

⇒ A = 22°

Question 6.

If A, B and C are interior angles of a triangle ABC, then show that sin\(\left(\frac{\mathbf{B}+\mathbf{C}}{2}\right)\) = cos\(\frac{A}{2}\)

Solution:

A, B and C are the interior angles of a triangle.

∴ A + B + C = 180°

[∵ The sum of the three angles of a triangle is 180°]

or B + C = 180° – A

or \(\frac{B+C}{2}\) = \(\frac{180^{\circ}-A}{2}\)

or \(\frac{B+C}{2}\) = (90° – \(\frac{\mathrm{A}}{2}\))

Taking sin on both sides

⇒ sin\(\left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)\) = sin(90° – \(\frac{\mathrm{A}}{2}\))

= cos\(\frac{\mathrm{A}}{2}\)

[∵ sin (90° – θ) = cos θ]

Hence Proved

Question 7.

Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.

Solution:

sin 67° + cos 75°

= sin (90° – 23°) + cos (90° – 15°)

= cos 23° + sin 15°

[∵ sin (90° – θ) = cos θ

and cos (90° – θ) = sin θ]

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