Question

if `3 cos theta = 1`, find the value of `(6 sin^2 theta + tan^2 theta)/(4 cos theta)`

Answer 1

Given `3 cos theta = 1`

We have to find the value of the expression `(6 sin^2 theta + tan^2 theta)/(4 cos theta)`

We have

`3 cos theta = 1`

`=> cos theta = 1/3`

`sin theta = sqrt(1 - cos^2 theta) =  sqrt(1- (1/3)^3) = sqrt8/3`

`tan theta = sin theta/cos theta = (sqrt8/3)/(1/3) = sqrt8`

Therefore,

`(6 sin^2 theta + tan^2 theta)/(4 cos theta) = (6 xx (sqrt8/3)^2 + (sqrt8)^2)/(4 xx 1/3)`

= 10

Hence, the value of the expression is 10.