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

Question

9066 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-15T07:46:45

#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 · 2017-01-15T07:47:22

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

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/((sinx/cosx+cosx/sinx))#

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

= #2/(1/(sinxcosx)#

= #2xx(sinxcosx)/1#

= #2sinxcosx#

= #sin2x#