1 cos ⁡290^ + 1 3 sin ⁡250^ =

Question

Accepted Answer

Correct answer: B. (b) Since 1 cos ⁡290^ + 1 3 sin ⁡250^ = 1 cos ⁡(270+20^ )+ 1 3 sin ⁡(270-20^ ) = 1 sin ⁡20^ - 1 3 sin ⁡20^ = 2( 32 cos ⁡20^ - 12 sin ⁡20^ ) 32 sin ⁡40^ = 2 sin ⁡(60^ -20^ ) 32 sin ⁡40^ = 4 3

$\frac{1}{cos 290 ^{\circ}} + \frac{1}{3 sin 250 ^{\circ}} =$

Solution

More Trigonometric Ratios & Identities PYQs