Question

Prove the identity. tanx-cotx/tanx+cotx=sin^2x-cos^2x Have to show the statements and the rules?

Answer 1

See below.

Explanation:

Identities:

`color(red)bb(tanx=sinx/cosx)`

`color(red)bb(cotx=cosx/sinx)`

Substituting these in the LHS:

`(sinx/cosx-cosx/sinx)/(sinx/cosx+cosx/sinx)`

Adding the terms in numerator and denominator:

`((sin^2x-cos^2x)/(sinxcosx))/((sin^2x+cos^2x)/(cosxsinx)`

Multiplying by `(cosxsinx)`

`(sin^2x-cos^2x)/(sin^2x+cos^2x)`

Identity:

`color(red)bb(sin^2x+cos^2x=1)`

Substituting this in the denominator:

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

`sin^2x-cos^2x`

As required:

`LHS-=RHS`

Answer 2

`(tanx-cotx)/(tanx+cotx)=sin^2x-cos^2x`

`=>(tanx-1/tanx)/(tanx+1/tanx)`

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

`=>(tan^2x-1)/(sec^2x)` `color(white)(wwwwww` `["as "tan^2x-sec^2x = 1]`

`=>(tan^2x/sec^2x-1/sec^2x)`

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