Tan^2x-sin^2x prove it can be written as tan^2xsin^2x?

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Answer 1 · 2017-11-13T15:37:52

See the proof below

We need

#tan^2x+1=sec^2x#

#(a+b)(a-b)=a^2-b^2#

Therefore,

#LHS=tan^2x-sin^2x=(tanx+sinx)(tanx-sinx)#

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

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

#=sin^2x(secx+1)(secx-1)#

#=sin^2x(sec^2x-1)#

#=sin^2xtan^2x#

#=RHS#

#QED#