HBSE 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3

Haryana State Board HBSE 10th Class Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Exercise Questions and Answers.

Haryana Board 10th Class Maths Solutions 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:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
= \(\frac{\sin \left(90^{\circ}-72^{\circ}\right)}{\cos 72^{\circ}}\)
= \(\frac{\cos 72^{\circ}}{\cos 72^{\circ}}\)
[∵ sin (90 – θ) = cos θ]
= 1
Hence, \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\) = 1.

(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}=\frac{\tan \left(90^{\circ}-64^{\circ}\right)}{\cot 64^{\circ}}\)
= \(\frac{\cot 64^{\circ}}{\cot 64^{\circ}}\)
[∵ sin (90 – θ) = cos θ]
= 1
Hence, \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\) = 1.

(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
= sin 42° – sin 42°
[∵ cos (90° – θ) = sin θ]
= 0.
Hence, cos 48° – sin 42° = 0.

(iv) cosec 31° – sec 59°
= cosec (90° – 59°) – sec 59°
= sec 59° – sec 59°
[∵ cosec (90° – θ) = sec θ]
= 0
Hence, cosec 31° – 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) We have,
L.H.S.= tan 48° tan 23° tan 42° tan 67°
= (tan 48° tan 42°) (tan 23° tan 67°)
= [tan (90° – 42°) tan 42°] [tan (90° – 67°) tan 67°]
= (cot 42° tan 42°) (cot 67° tan 67°)
[∵ tan (90 – θ) = cot θ]
= (cot 42° . \(\frac{1}{\cot 42^{\circ}}\)) (cot 67° . \(\frac{1}{\cot 67^{\circ}}\))
= 1 × 1 = 1 = RHS
∴ L.H.S. = R.H.S.
Hence Proved.

(ii) We have,
L.H.S. = cos 38° cos 52° – sih 38° sin 52°
= cos (90° – 52°) cos 52° – sin (90° – 52°) sin 52°
= sin 52° cos 52° – cos 52° sin 52°
= cos 52° sin 52° – cos 52° sin 52°
= 0 = R.H.S.
∴ L.H.S. = R.H.S.
Hence Proved.

Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution :
We have,
tan 2A = cot (A – 18°)
⇒ cot (90° – 2A) = cot (A – 18°)
[∵ tan θ = cot (90° – θ)]
⇒ 90° – 2A= A – 18°
⇒ 90° + 18° = A + 2A
⇒ 3A = 108°
A = \(\frac{108^{\circ}}{3}\) = 36°.
Hence, A = 36°.

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

Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
We have,
sec 4A = cosec (A – 20°)
sec 4A = sec [90° – (A – 20°)]
[∵ cosec θ = sec (90° – θ)]
4A = 90° – (A – 20°)
4A = 90° – A + 20°
5A = 110°
A = \(\frac{110^{\circ}}{5}\) = 22°.
Hence, A = 22°.

Question 6.
If A, B and C are interior angles of a triangle ABC, then show that :
sin \(\left(\frac{B+C}{2}\right)\) = cos \(\frac{\mathrm{A}}{2}\).
Solution :
We know that the sum of angles of a triangle is 180°.
∴ A + B + C = 180°
⇒ B + C = 180° – A
\(\frac{\mathrm{B}+\mathrm{C}}{2}=\frac{180^{\circ}-\mathrm{A}}{2}\)
\(\frac{\mathrm{B}+\mathrm{C}}{2}=90^{\circ}-\frac{\mathrm{A}}{2}\)
\(\sin \left(\frac{\mathrm{B}+\mathrm{C}}{2}\right)=\sin \left(90^{\circ}-\frac{\mathrm{A}}{2}\right)\)
\(\sin \left(\frac{B+C}{2}\right)=\cos \frac{A}{2}\)
[∵ sin (90° – θ) = cos θ]
Hence Proved.

Question 7.
Express sin 67° + cos 75° in terms of trigonometric ratios of singles between 0° and 45°.
Solution :
We have,
sin 67° + cos 75° = sin (90° – 23°) + cos (90 – 15°) = cos 23° + sin 15°
[∵ sin(90 – θ) = cos θ
cos (90 – θ) = sin θ]
Hence, sin 67° + cos 75° = cos 23° + sin 15°.

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