Question

Question #be269

Answer 1

See proof below

Explanation:

We need

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

`cscx=1/sinx`

`cotx=cosx/sinx`

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

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

Therefore,

`LHS=(cscx-1)(sinx+1)`

`=(1/sinx-1)(1+sinx)`

`=((1-sinx)(1+sinx))/sinx`

`=(1-sin^2x)/sinx`

`=cos^2x/sinx`

`=cosx*cosx/sinx`

`=cosxcotx`

`=RHS`

`QED`