Prove that $(sinx-1)/(sinx+1)=(-cos^2x)/(sinx+1)^2$ ?

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Answer 1 · 2017-01-15T08:05:44

$LHS=(sinx-1)/(sinx+1)$

$=((sinx-1)(sinx+1))/((sinx+1)(sinx+1))$

$=(sin^2x-1)/(sinx+1)^2$

$=(-(cos^2x))/(sinx+1)^2=RHS$

Proved

Answer 2 · 2017-01-15T08:06:54

Please see below.

$(sinx-1)/(sinx+1)$

= $((sinx-1)(sinx+1))/((sinx+1)(sinx+1))$

= $(sin^2x-1)/(sinx+1)^2$

= $(-(1-sin^2x))/(sinx+1)^2$

= $(-cos^2x)/(sinx+1)^2$

Prove that (sinx-1)/(sinx+1)=(-cos^2x)/(sinx+1)^2(sinx-1)/(sinx+1)=(-cos^2x)/(sinx+1)^2?