Question #0f589

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Answer 1 · 2016-12-26T07:59:53

$RHS=1-sin^2x/(1+cotx)-cos^2x/(1+tanx)$

$=1-sin^3x/(sinx+sinxcotx)-cos^3x/(cosx+cosxtanx)$

$=1-(sin^3x/(sinx+cosx)+cos^3x/(cosx+sinx))$

$=1-((sin^3x+cos^3x)/(cosx+sinx))$

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

$=1-(sin^2x-sinxcosx+cos^2x)$

$=1-(1-sinxcosx)$

$=sinxcosx$

But $LHS=1-sinxcosx$

Hence $LGS!=RHS$

So it must be an equation which is

$1-sinxcosx=sinxcosx$

$=>2sinxcosx=1$

$=>sin2x=sin(pi/2)$

$=>2x=npi+(-1)^n(pi/2)$

$=>x=(npi)/2+(-1)^n(pi/4)" where "n in ZZ$