# KSEEB SSLC Class 10 Maths Solutions Chapter 11 Introduction to Trigonometry Ex 11.3

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### Karnataka State Syllabus Class 10 Maths SolutionsChapter 11 Introduction to Trigonometry Ex 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 = 108∘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 = 110∘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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