Question

If cosec θ + cot θ = p, then prove that cos θ = `(p^2 - 1)/(p^2 + 1)`

Answer 1

According to the question,

cosec θ + cot θ = p

Since, cosec θ = `1/sintheta` and cot θ = `costheta/sintheta`

`1/sintheta + costheta/sintheta` = p

`(1 + costheta)/sintheta` = p

Squaring on L.H.S and R.H.S,

`((1 + costheta)/sin theta)^2` = p²

`(1 + cos^2 theta + 2 cos theta)/(sin^2 theta)` = p²

Applying component and dividend rule,

`((1 + cos^2 theta + 2 cos theta) - sin^2 theta)/((1 + cos^2 theta + 2 cos theta) + sin^2 theta) = ("p"^2 - 1)/("p"^2 + 1)`

= `((1 - sin^2theta) + cos^2 theta + 2 cos theta)/(sin^2 theta + cos^2 theta + 1 + 2 cos theta) = ("p"^2 - 1)/("p"^2 + 1)`

Since, 1 – sin²θ = cos²θ and sin²θ + cos²θ = 1

`(cos^2 theta + cos^2 theta + 2 cos theta)/(1 + 1 + 2 cos theta) = ("p"^2 - 1)/("p"^2 + 1)`

`(2 cos^2 theta + 2 cos theta)/(2 + 2 cos theta) = ("p"^2 - 1)/("p"^2 + 1)`

`(2 cos theta(cos theta + 1))/(2(cos theta + 1)) = ("p"^2 - 1)/("p"^2 + 1)`

cos θ = `("p"^2 - 1)/("p"^2 + 1)`

Hence proved.