If sin x : cos x : : 3 : 1 , find sin x , cos x .

Given that, sin x : cos x : : 3 : 1 ⇒ sin x cos x = 3 ...(i) ⇒ tan x = 3 ⇒ x lies in I st or III rd quadrant. We know that, sec 2 x - tan 2 x = 1 ⇒ sec x = 1 + 3 ⇒ sec x = | 2 | ⇒ sec x = ± 2 [since sec x is positive in the first quadrant and negative in the third quadrant] ⇒ cos x = 1 sec x = ± 1 2 [Since cos x is positive in the first quadrant and negative in the third quadrant] From (i), sin x = 3 cos x ⇒ sin x = ± 3 2 [Since sin x is positive in the first quadrant and negative in the third quadrant] Hence, sin x = ± 3 2 and cos x = ± 1 2

If cos x = - 3 5 and π < x < 3 π 2 , find the other t - ratios and hence evaluate cosec x + cot x sec x - tan x .

Given that, cos x = - 3 5 and π < x < 3 π 2 ⇒ x lies in III rd quadrant. In III r d quadrant, only tan x and cot x are positive and rest are negative. We know that, sin 2 x + cos 2 x = 1 ⇒ sin x = 1 - 9 25 = 16 25 ⇒ sin x = - 4 5 [ ∵ x ∈ III r d quadrant ] cosec x = 1 sin x = - 5 4 sec x = 1 cos x = - 5 3 We know that, sec 2 x - tan 2 x = 1 ⇒ tan x = 25 9 - 1 = 16 9 ⇒ tan x = 4 3 cot x = 1 tan x = 3 4 Now, cosec x + cot x sec x - tan x = - 5 4 + 3 4 - 5 3 - 4 3 = - 2 4 - 9 3 = - 1 2 × - 1 3 ∴ cosec x + cot x sec x - tan x = 1 6 .

If cosec x - cot x = 3 2 , find cos x . In which quadrant does x lie?

Given that, cosec x - cot x = 3 2 .....(i) We know that, cosec 2 x - cot 2 x = 1 ⇒ ( cosec x - cot x ) ( cosec x + cot x ) = 1 ⇒ ( cosec x + cot x ) = 1 3 2 ⇒ ( cosec x + cot x ) = 2 3 ...(ii) (i)+(ii) ⇒ 2 cosec x = 3 2 + 2 3 ⇒ cosec x = 1 2 9 + 4 6 ⇒ cos e c x = 13 12 From (i), cot x = 13 12 - 3 2 ⇒ cot x = 13 - 18 12 ⇒ cot x = - 5 12 We got cosec x as positive and cot x as negative, implies x lies in II quadrant. Now, sin x = 1 cosec x = 12 13 We know that, sin 2 x + cos 2 x = 1 ⇒ cos x = 1 - 144 169 ⇒ cos x = 12 13 ⇒ cos x = - 12 13 ∵ x ∈ II n d quadrant Hence, cos x = - 12 13 and x lies in the second quadrant.

Show that sin π 6 cos 0 + sin π 4 cos π 4 + sin π 3 cos π 6 = 7 4 .

To Prove : sin π 6 cos 0 + sin π 4 cos π 4 + sin π 3 cos π 6 = 7 4 LHS = sin π 6 cos 0 + sin π 4 cos π 4 + sin π 3 cos π 6 = 1 2 · 1 + 1 2 · 1 2 + 3 2 · 3 2 = 1 2 + 1 2 + 3 4 = 7 4 = RHS Hence, proved.

Show that sin 2 π 6 + cos 2 π 3 - tan 2 π 4 = - 1 2 .

To Prove : sin 2 π 6 + cos 2 π 3 - tan 2 π 4 = - 1 2 LHS = sin 2 π 6 + cos 2 π 3 - tan 2 π 4 = 1 2 2 + 1 2 2 - 1 2 = 1 4 + 1 4 - 1 = - 1 2 = RHS Hence, proved.

Show that 4 sin π 6 sin 2 π 3 + 3 cos π 3 tan π 4 + cosec 2 π 2 = 2 sec 2 π 4 .

To Prove : 4 sin π 6 sin 2 π 3 + 3 cos π 3 tan π 4 + cosec 2 π 2 = 2 sec 2 π 4 LHS = 4 sin π 6 sin 2 π 3 + 3 cos π 3 tan π 4 + cosec 2 π 2 = 4 1 2 3 2 2 + 3 1 2 · 1 + 1 2 = 3 2 + 3 2 + 1 = 4 ...(i) Now, RHS = 2 sec 2 π 4 = 2 ( 2 ) 2 = 2 × 2 = 4 ...(ii) From (i) and (ii) 4 sin π 6 sin 2 π 3 + 3 cos π 3 tan π 4 + cosec 2 π 2 = 2 sec 2 π 4 Hence, proved.