Question

Simplify `[2(tanx-cotx)]/(tan^2x-cot^2x)`?

Answer 1

`LHS=(2(tanx-cotx))/(tan^2x-cot^2x)`

`=(2(tanx-cotx))/((tanx-cotx)(tanx+cotx))`

`=(2tanx)/(tanx(tanx+cotx))`

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

Proved

Answer 2

`[2(tanx-cotx)]/(tan^2x-cot^2x)=sin2x`

Explanation:

Using the identity `a^2-b^2=(a+b)(a-b)`,

we can split the term `(tan^2x-cot^2x)=(tanx+cotx)(tanx-cotx)` and we get

`[2(tanx-cotx)]/(tan^2x-cot^2x)`

= `[2(tanx-cotx)]/((tanx+cotx)(tanx-cotx))`

= `2/((tanx+cotx))`

= `2/((tanx+cotx))`

= `2/((sinx/cosx+cosx/sinx))`

= `2/((sin^2x+cos^2x)/(sinxcosx)`

= `2/(1/(sinxcosx)`

= `2xx(sinxcosx)/1`

= `2sinxcosx`

= `sin2x`