Prove this identity?

Question

#(1+cosx)/(1-cosx)# = #(tan^2x)/(secx-1)^2#

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Answer 1 · 2018-07-15T18:23:04

Consider

#LHS=\frac{1+\cos x}{1-\cos x}#

#=\frac{1+\1/sec x}{1-1/\sec x}#

#=\frac{sec x+1}{\sec x-1}#

#=\frac{(sec x+1)(\sec x+1)}{(\sec x-1)(\sec x+1)}#

#=\frac{(sec x+1)^2}{\sec^2 x-1}#

#=\frac{(sec x+1)^2}{\tan^2 x}#

#RHS=\frac{tan^2x}{(\sec^2 x-1)^2}#

#=\frac{tan^2x}{(\tan^2 x)^2}#

#=\frac{tan^2x}{\tan^4 x}#

#=\frac{1}{\tan^2 x}#

#=cot^2 x#

#LHS\ne RHS#

Answer 2 · 2018-07-15T18:24:58

Please see below.

We know that ,

#color(red)((1)sec^2theta-tan^2theta=1#

Here ,

#LHS=(1+cosx)/(1-cosx)#

#color(white)(LHS)=(1+1/secx)/(1-1/secx)#

#color(white)(LHS)=(secx+1)/(secx-1)#

Multiplying numerator and denominator by : #color(blue)( (secx-1)#

#LHS=(secx+1)/(secx-1) *color(blue)( (secx-1)/(secx-1)#

#color(white)(LHS)=(sec^2x-1)/((secx-1)^2)tocolor(red)(Apply(1)#

#color(white)(LHS)=tan^2x/(secx-1)^2#

#LHS=RHS#