Question

Question #a32b3

Answer 1

See proof below

Explanation:

We use

`(a+b)^2=a^2+2ab+b^2`

`sin^2x+cos^2x=1`

Dividing by `cos^2x`

`sin^2x/cos^2x+1=1/cos^2x`

`tan^2x+1=sec^2x`

`tan^2x=sec^2x-1`

Therefore,

`LHS=(tanx+secx)^2`

`=tan^2x+sec^2x+2tanxcosx`

`=sec^2x-1+sec^2x+2tanxsecx`

`=2sec^2x+2tanxsecx-1`

`=RHS`

`QED`