NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

These NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Exercise 8.3

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 1.
(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:
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 1

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) 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° = tan 48° tan 23° .\(\frac{1}{\tan 48^{\circ}} \cdot \frac{1}{\tan 23^{\circ}}\)
= 1 = RHS [Verified]

(ii) LHS = 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° = 0 = RHS [Verified]

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

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°) [∵cot (90° – θ) = tan θ]
⇒ 90° – 2A = A – 18° ⇒ 3A = 108° ⇒ A = \(\frac { 108° }{ 3 }\)
∴ ∠ A = 36°

Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
tan A = cot B ⇒ tan A = tan (90° – B) [ ∵ tan (90° – θ) = cot θ]
⇒ A = 90° – B ⇒ A + B = 90° Proved

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°) [cosec (90° – θ) = sec θ]
⇒ 90° – 4A = A – 20° ⇒ 5A = 110°
A = \(\frac { 110° }{ 5 }\)
A = 22°
∴ ∠ A = 22°

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 6.
If A, Band Care interior angles of a triangle ABC, then show that: sin (\(\frac { B+C }{ 2 }\)) = cos \(\frac { A }{ 2 }\)
Solution:
We have given that,
A, B and C are the interior angles of a triangle ABC
NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3 2

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry Ex 8.3

Question 7.
Express sin 61° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution:
sin 67° + cos 75°
or sin (90° – 23°) + cos (90° – 15°)
or cos 23° + sin 15° (∵ sin (90° – θ) = cos θ
and cos (90° – θ) = sin θ)
Now, sin 67° + cos 75° in term if trigonometric ratios of angles between 0° and 45°.
(cos 90° – 67°) + (sin 90° – 75°)
= cos 23° + sin 15°

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