Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 8.3 Textbook Questions and Answers.
BSEB Bihar Board Class 10th Maths Solutions Chapter 8 Introduction to Trigonometry Ex 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:
(i) \(\frac{\sin 18^{\circ}}{\cos 72^{\circ}}\)
= \(\frac{\sin \left(90^{\circ}-72^{\circ}\right)}{\cos 72^{\circ}}\)
= \(\frac{\cos 72^{\circ}}{\cos 72^{\circ}}\)
= 1
(ii) \(\frac{\tan 26^{\circ}}{\cot 64^{\circ}}\)
= \(\frac{\tan \left(90^{\circ}-64^{\circ}\right)}{\cot 64^{\circ}}\)
= \(\frac{\cot 64^{\circ}}{\cot 64^{\circ}}\)
= 1
(iii) cos 48° – sin 42° = cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59° = cosec (90° – 59°) – sec 59°
= sec 59° – 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) tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
= cot 42° cot 67° tan 42° tan 67°
= \(\frac{1}{\tan 42^{\circ}}\).\(\frac{1}{\tan 67^{\circ}}\).tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52°
= cos (90° – 52°) cos (90° – 38°) – sin 38° sin 52°
= sin 52° sin 38° – sin 38° sin 52° = 0
Question 3.
If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
We are given that
tan 2A = cot (A – 18°) … (1)
Since tan 2A = cot (90° – 2A), so we can write (1) as
cot (90° – 2A) = cot (A – 18°)
Since (90° – 2A) and (A – 18°) are both acute angles, therefore
90° – 2A = A – 18°
or – 2A – A = – 18° – 90° or – 3A = – 108°
or A = 36°
Question 4.
If tan A = cot B, prove that A + B = 90°.
Solution:
We are given that
tan A = cot B … (1)
Since tan A = cot (90° – A), so we can write (1)
as cot (90° – A) = cot B
Since (90° – A) and B are both acute angles, therefore
90° – A = B or A + B = 90°
Question 5.
If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
We are given that
sec 4A = cosec (A – 20°). …(1)
Since sec 4A = cosec (90° – 4A), so we can write (1) as
cosec (90° – 4A) = cosec (A – 20°)
Since (90° – 4A) and (A – 20°) are both acute angles, therefore
90° – 4A = A – 20° or – 4A – A = – 20° – 90°
or – 5A = – 110° or A = 22°
Question 6.
If A, B and C are, interior angles of a ∆ ABC, then show, that
sin\(\left(\frac{B+C}{2}\right)\) = cos\(\frac { A }{ 2 }\)
Solution:
Since A, B and C are the interior angles of a ∆ ABC, therefore
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°