Question

Prove the following:

`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ

Answer 1

L.H.S = `sintheta/(1 + cos theta) + (1 + cos theta)/sintheta`

Taking the L.C.M of the denominators,

We get,

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

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

Since, sin²θ + cos²θ = 1

= `(1 + 1 + 2costheta)/((1 + costheta) * sin theta)`

= `(2 + 2 cos theta)/((1 + cos theta) * sin theta)`

= `(2(1 + cos theta))/((1 + cos theta) * sin theta)`

Since, `1/sin theta` = cosec θ

= `2/sin theta`

= 2 cosec θ

R.H.S

Hence proved.