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 48^o - sin 42^o

Solution: cos 48^o - sin 42^o

= cos (90^o - 42^o) – sin 42^o

= sin 42^o - sin 42^o = 0

(iv) cosec 31^o - sec 59^o

Solution: cosec 31^o - sec 59^o

= cosec(90^o - 59^o) – sec 59^o

= sec 59^o - sec 59^o = 0

Question – 2 - Show that:

(i) tan 48^o tan 23^o tan 42^o tan 67^o = 1

Solution: LHS: tan 48^o tan 23^o tan 42^o tan 67^o

= tan (90^o - 42^o) x tan 23^o x tan 42^o x tan (90^o - 23^o)

= cot 42^o x tan 23^o x tan 42^o x cot 23^o

= cot 42^o x tan 42^o x tan 23^o x cot 23^o

= 1 x 1 = 1 Proved

(ii) cos 38^o cos 52^o - sin 38^o sin 52^o = 0

Solution: cos 38^o cos 52^o - sin 38^o sin 52^o = 0

= cos (90^o - 52^o) cos (90^o - 38^o) - sin 38^o sin 52^o

= sin 38^o sin 52^o - sin 38^o sin 52^o = 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°