Question

Find the value of x for which `2 "cosec"^2 30^circ + x sin^2 60^circ - 3/4 tan^2 30^circ = 10`.

Answer 1

Given: `2 "cosec"^2 30^circ + x sin^2 60^circ - 3/4 tan^2 30^circ = 10`

`"cosec"  30^circ = 1/(sin 30^circ) = 2`

⇒ `"cosec"^2 30^circ = 4`

⇒ 2·cosec² 30° = 8

`sin 60^circ = sqrt(3)/2`

⇒ `sin^2 60^circ = 3/4`

So `x·sin^2 60^circ = x · (3/4)`.

`tan 30^circ = 1/sqrt(3)`

⇒ `tan^2 30^circ = 1/3`

⇒ `(3/4)·tan^2 30^circ = (3/4)·(1/3) = 1/4`.

Substitute into (ii): `8 + (3/4)x - 1/4 = 10`.

Combine constants: `8 - 1/4 = 31/4`.

So `(3/4)x + 31/4 = 10 = 40/4`.

`(3/4)x = 40/4 - 31/4 = 9/4`

⇒ Multiply by 4

⇒ 3x = 9

⇒ x = 3