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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