In this article, we will share MP Board Class 10th Maths Book Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Pdf, These solutions are solved subject experts from the latest edition books.

## MP Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.

Evaluate:

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

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

(iii) cos48° – sin42°

(iv) cosec31°- sec59°

Solution:

(iii) cos 48° – sin 42°

cos 48° = cos (90° – 42°) = sin 42°

[∵ cos (90° – A) = sin A]

∴ cos 48° – sin 42° = sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

cosec 31° = cosec (90° – 59°) = sec 59° [ ∵ cosec (90° – A) = sec A]

∴ cosec 31° – sec 59° = sec 59° – sec 59° = 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° tan 42° tan 67°

= tan (90° – 42°) tan 23° tan 42° tan (90° – 23°)

= cot 42° tan 23° tan 42° cot 23° [ ∵ tan (90° – A) = cot A]

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

= cos 38° cos (90° – 38°) – sin38° sin(90° – 38°)

= cos 38° sin 38° – sin 38° cos 38° [ ∵ sin(90° – A) = cosA and cos(90° – A)= sinA]

= 0 = R.H.S.

⇒ 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:

tan 2A = cot(A – 18°)

cot(90 – 2A) = cot (A – 18)

90 – 2A = A – 18

90 + 18 = A + 2A

3A= 108

A = \(\frac {108}{3}\)

A = 36°.

Question 4.

If tan A = cotB, prove that A + B = 90°.

Solution:

tan A = cot B and cot B = tan (90° – B) [∵ tan (90° – θ) = cot θ]

∴ 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:

sec 4A= cosec (A – 20)

cosec (90 – 4A) = cosec (A – 20)

90 – 4A = A – 20

90 + 20 =A + 4A

110 = 5A

A = \(\frac{110}{5}\)

A = 22°.

Question 6.

If A, 6 and C are interior angles of a triangle ABC, then show that \(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)

Solution:

Since, sum of the angles of ∆ABC is 180° i.e.,

A + B + C = 180°

∴ B + C = 180° – A

Dividing both sides by 2, we get

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)

[sin(90 – θ) = cosθ

cos(90 – θ) = sinθ]

= cos 23 + sin 15