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Answer 1 · 2018-02-06T18:18:21

Please refer to a Proof in the Explanation.

#tan2x=tan(x+x)#,

#=(tanx+tanx)/(1-tanx*tanx)#,

#rArr tan2x=(2tanx)/(1-tan^2x)...............................................(star_1)#.

On the other hand, #tan2x=(sin2x)/(cos2x)#,

#=(2sinxcosx)/(cos^2x-sin^2x)#,

#=(2sinxcosx)/{(1-sin^2x)-sin^2x}#.

# rArr tan2x=(2sinxcosx)/(1-2sin^2x).......................................(star_2)#.

Hence, by #(ast_1) and (ast_2)#,

# (cancel(2)tanx)/(1-tan^2x)=(cancel(2)sinxcosx)/(1-2sin^2x), or, #

# (sinxcosx)/(1-2sin^2x)=tanx/(1-tan^2x)#,

#=cancel(tanx)/{cancel(tanx)(1/tanx-tanx)}#.

#rArr (sinxcosx)/(1-2sin^2x)=1/(cotx-tanx)#.