Introduction to Trigonometry Ex 8.3 Solutions
Question - 1 : - Evaluate :
Answer - 1 : -
(i) sin 18°/cos 72° (ii) tan 26°/cot 64°
(iii) cos48° – sin 42 (iv) cosec31° – 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
Answer - 2 : - 
Question - 3 : - If tan 2A = cot (A –18°), where 2A is an acute angle, find the value of A.
Answer - 3 : - 
Question - 4 : - If tan A = cot B,prove that A + B = 90°.
Answer - 4 : - 
Question - 5 : - If sec 4A = cosec (A– 20°), where 4A is an acute angle, find the value of A.
Answer - 5 : - 
Question - 6 : - If A, B and C areinterior angles of a triangle ABC, then show that
sin(B+C/2) = cos A/2
Answer - 6 : -
We know that, for a given triangle, sum of allthe interior angles of a triangle is equal to 180°
A + B + C = 180° ….(1)
To find the value of (B+ C)/2, simplify theequation (1)
⇒ B + C = 180° – A
⇒ (B+C)/2 = (180°-A)/2
⇒ (B+C)/2 = (90°-A/2)
Now, multiply both sides by sin functions, weget
⇒ sin (B+C)/2 = sin (90°-A/2)
Since sin (90°-A/2) = = cos A/2, the aboveequation is equal to
sin (B+C)/2 = cos A/2
Hence proved.
Question - 7 : - Express sin 67° +cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Answer - 7 : -
Given:
sin 67° + cos 75°
In term of sin as cos function and cos as sinfunction, it can be written as follows
sin 67° = sin (90° – 23°)
cos 75° = cos (90° – 15°)
= sin (90° – 23°) + cos (90° – 15°)
Now, simplify the above equation
= cos 23° + sin 15°
Therefore, sin 67° + cos 75° is also expressedas cos 23° + sin 15°