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#