**KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.3.

## Karnataka SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Exercise 11.3

Question 1.

Evaluate :

iii) cos 48° – sin 42°

iv) cosec 31° – sec 59°

Solution:

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) tan 48° tan 23° tan 42° tan 67° = 1

LHS = tan 48° tan 23° tan 42° tan 67°

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

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

= 1

∴ LHS = RHS

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

cos 38°. cos 52 – sin 38° sin 52

= cos 38° . cos52 – sin(90 – 52°) sin (90 – 38°)

= cos38° . cos52 – cos52 cos32°

= 0

Question 3.

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

Solution:

Since, tan 1A = cot (A – 18°)

Also, tan (2A) = cot (90° – 2A) [∵ tan θ = cot (90° – θ)]

∴ A – 18° = 90° – 2A

⇒ A + 2A = 90° + 18°

⇒ 3A = 108° ⇒ A = \(\frac{108^{\circ}}{3}\) = 36°

Question 4.

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

Solution:

tan A = cot B

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:

sec 4A = cosec (A – 20°)

Also, sec 4A = cosec (90° – 4A) [ ∵ cosec (90° – θ) = sec θ]

∴ A – 20° = 90° – 4A

⇒ A + 4A = 90° + 20°

⇒ 5A = 110° ⇒ A = \(\frac{110^{\circ}}{5}\) = 22°

Question 6.

If A, B, and C are interior angles of a triangle ABC, then show that

Solution:

A + B + C = 180°

B + C = 180 – A

Question 7.

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

Solution:

Since, sin 67° = sin (90° – 23°) = cos 23° [ ∵ sin (90° – θ ) = cos θ]

Also, cos 75° = cos (90° – 15°) = sin 15° [∵ cos (90° – θ) = sin θ]

∴ sin 67° + cos 75° = cos 23° + sin 15°

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