Question

Question #4c761

Answer 1

`LHS=2tanxcsc2x-tan^2x`

`=(2tanx)1/(sin2x)-tan^2x`

`=(2tanx)1/((2tanx)/(1+tan^2x))-tan^2x`

`=((2tanx)(1+tan^2x))/(2tanx)-tan^2x`

`=1+tan^2x-tan^2x=1=RHS`

Verified

Answer 2

Use the following identities:

`•tanx = sinx/cosx`
`•cscx = 1/sinx`
`•sin2x = 2sinxcosx`
`•cos^2x + sin^2x = 1`

Using these identities gradually, we have

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

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

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

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

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

`1 = 1`

This is obviously true, so the identity has been proved.

Hopefully this helps!