# Prove the identity: `tanx/(secx+1) = (secx-1)/tanx`

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LS= tanx/secx+1

=tanx(secx-1)/secx+1(secx-1)

=tanx(secx-1)/sec^2x-1

=tanx(secx-1)/tan^2x

=secx-1/tanx

RS= secx-1/tanx

Therefore, LS=RS

The identity `tan x/(sec x + 1) = (sec x - 1)/tan x` has to be proved.

`tan x/(sec x + 1)`

=> `(sin x/cos x)/(1/cos x + 1)`

=> `(sin x)/(1 + cos x)`

=> `((sin x)(1 - cos x))/((1 + cos x)(1 - cos x))`

=> `((sin x)(1 - cos x))/(1 - cos^2x)`

=> `((sin x)(1 - cos x))/(sin^2x)`

=> `(1 - cos x)/sin x`

=> `(1/cos x - 1)/tan x`

=> `(sec x - 1)/tan x`

**This proves that **`tan x/(sec x + 1) = (sec x - 1)/tan x`

It is required to prove that: By cross multipling ;

tan^2x = sec^2-1

and this is a basic idenitity

Q.E.D

LS= tanx/secx+1

=tanx(secx-1)/secx+1(secx-1) =tanx(secx-1)/sec^2x-1

=tanx(secx-1)/tan^2x =secx-1/tanx

RS= secx-1/tanx

Therefore, LS=RS

Hence, proven.