Question

`((2 tan^2 30^circ sec^2 52^circ sin^2 38^circ)/("cosec"^2 70^circ - tan^2 20^circ))` = ?

Options

• 2

• 1

• `2/3`

• `3/2`

Answer 1

`bb(2/3)`

Explanation:

Given expression = `{(2 xx (1/sqrt(3))^2 sec^2 52^circ sin^2 (90^circ - 52^circ))/("cosec"^2 70^circ - tan^2 (90^circ - 70^circ))}`

= `2/3 xx ((sec^2 52^circ cos^2 52^circ)/("cosec"^2 70^circ - cot^2 70^circ))`

= `2/3 xx 1/1`

= `2/3`