Trigonometry
NCERT Exercise 8.3
Question – 1 - Evaluate the following:
(i) `(text(sin)18°)/(text(cos)72°)`
Solution:
`(text(sin)18°)/(text(cos)72°)`
`=(text(sin)(90°-72°))/(text(cos)72°)`
`=(text(cos)72°)/(text(cos)72°)=1`
(ii) `(text(tan)26°)/(text(cot)64°)`
Solution:
`(text(tan)26°)/(text(cot)64°)`
`=(text(tan)(90°-64°))/(text(cot)64°)`
`=(text(cot)64°)/(text(cot)64°)=1`
(iii) cos 48o - sin 42o
Solution: cos 48o - sin 42o
= cos (90o - 42o) – sin 42o
= sin 42o - sin 42o = 0
(iv) cosec 31o - sec 59o
Solution: cosec 31o - sec 59o
= cosec(90o - 59o) – sec 59o
= sec 59o - sec 59o = 0
Question – 2 - Show that:
(i) tan 48o tan 23o tan 42o tan 67o = 1
Solution: LHS: tan 48o tan 23o tan 42o tan 67o
= tan (90o - 42o) x tan 23o x tan 42o x tan (90o - 23o)
= cot 42o x tan 23o x tan 42o x cot 23o
= cot 42o x tan 42o x tan 23o x cot 23o
= 1 x 1 = 1 Proved
(ii) cos 38o cos 52o - sin 38o sin 52o = 0
Solution: cos 38o cos 52o - sin 38o sin 52o = 0
= cos (90o - 52o) cos (90o - 38o) - sin 38o sin 52o
= sin 38o sin 52o - sin 38o sin 52o = 0 proved
Question – 3 - If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution: tan 2A = cot (90° – 2A)
This means;
90° – 2A = A – 18°
Or, 108° – 2A = A
Or, 3A = 108°
Or, A = 108/3 = 36°
Question – 4 - If tan A = cot B, prove that A + B = 90°
Solution: tan A = cot (90°- A) = cot B
This means, 90° - A = B
Or, 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 (90° - 4A) = cosec (A - 20°)
This means; 90° – 4A = A - 20°
Or, 110°– 4A = A
Or, 5A = 110°
Or, A = 22°
Question – 6 - If A, B and C are interior angles of a triangle ABC, then show that
`text(sin)((B+C)/(2))=text(cos)A/2`
Solution: Since, A + B + C = 90°
So, B + C = 90°- A
Hence,
`text(sin)((B+C)/(2))=text(sin)((90°-A)/(2))`
`=text(cos)A/2` proved
Question – 7 - Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Solution: sin 67° + cos 75° can be written as follows:
Sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°