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

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Answer 1 · 2016-11-26T16:45:34

Please see below.

$(tanx-cotx)/(tanx+cotx)$

= $(sinx/cosx-cosx/sinx)/(sinx/cosx+cosx/sinx)$ - (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)

= $sin^2x-cos^2x$

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