Question

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

Answer 1

Please see below.

Explanation:

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

= `(sinx/cosx-cosx/sinx)/(sinx/cosx+cosx/sinx)` - (converting tan ad cot ratios into sin and cos ratios)

= `((sin^2x-cos^2x)/(sinxcosx))/((sin^2x+cos^2x)/(sinxcosx))` - (adding as in fractions)

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

= `(sin^2x-cos^2x)/cancel(sinxcosx)xxcancel(sinxcosx)/1` - (as `sin^2x+cos^2x=1` and cancelling like terms in numerator and denominator)

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