Question

How do you verify `cot^2x*cos^2x = cot^2x-cos^2x`?

Answer 1

See the proof below

Explanation:

We need

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

`cotx=cosx/sinx`

Therefore,

`LHS=cot^2xcos^2x`

`=cot^2x(1-sin^2x)`

`=cot^2x-cot^2xsin^2x`

`=cot^2x-cos^2x/cancel(sin^2x)*cancel(sin^2x)`

`=cot^2x-cos^2x`

`=RHS`

`QED`

Answer 2

See the answer below...

Explanation:

`cot^2x cdot cos^2x`

`=cot^2x cdot (1-sin^2x)`

`=cot^2x-cot^2x cdot sin^2x`

`=cot^2x-cos^2x/sin^2x cdot sin^2x`

`=cot^2x-cos^2x" "`[Proved...]

Hope it helps...
Thank you...