# Prove the identity `cos^2 x . tan x = (2sinx)/(secx+cosx+sin^2 x. sec x)`

*print*Print*list*Cite

### 2 Answers

`cos^2 x . tanx = (2sinx)/(secx+cosx+sin^2 x . sec x)`

`cos^2 x . tanx`

`=> cos^2 x. (sinx/cosx) `

=cosx. sinx ----(1)

`(2.sinx)/(sec x+cosx+sin^2 x sec x)`

`(2.sinx)/((1/cos x) + (cosx) + (sin^2 x)/(cosx))`

`(2.sinx.cosx)/(1+cos^2 x+ sin^2 x)`

`(2. sinx . cosx)/(2)`

sinx.cosx--------(2)

As (1) = (2) So `cos^2 x . tanx = (2sinx)/(secx+cosx+sin^2 x . sec x)`

The identity `cos^2 x* tan x = (2sinx)/(secx+cosx+sin^2 x* sec x) ` has to be proved.

`(2sinx)/(secx+cosx+sin^2 x* sec x) `

= `(2sinx)/(1/(cos x)+cosx+sin^2 x*1/(cos x)) `

= `(2sinx*cos x)/(1+cos^2x+sin^2 x) `

= `(2sinx*cos x)/(1+1)`

= `(2sinx*cos x)/2`

= `sinx*cos x`

= `(sin x/cos x)*cos x*cos x`

= `tan x*cos^2x`

**This proves that **` cos^2 x* tan x = (2sinx)/(secx+cosx+sin^2 x* sec x)`