Question #ba02a

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Answer 1 · 2018-02-14T06:06:44

Prove that $(sin^3x+cos^3x)/(1-2cos^2x)=(secx-sinx)/(tanx-1)$

$RHS=(secx-sinx)/(tanx-1)$

$color(white)(RHS)=(1/cosx-sinx)/(sinx/cosx-1)$

$color(white)(RHS)=(1-sinxcosx)/(sinx-cosx)$

$color(white)(RHS)=(1-sinxcosx)/(sinx-cosx)*((sinx+cosx))/((sinx+cosx))$

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

$color(white)(RHS)=(sinx+cosx-sin^2xcosx-sinxcos^2x)/(1-cos^2x-cos^2x)$

$color(white)(RHS)=(sinx-sinxcos^2x+cosx-sin^2xcosx)/(1-2cos^2x)$

$color(white)(RHS)=(sinx(1-cos^2x)+cosx(1-sin^2x))/(1-2cos^2x)$

$color(white)(RHS)=(sinx*sin^2x+cosx*cos^2x)/(1-2cos^2x)$

$color(white)(RHS)=(sin^3x+cos^3x)/(1-2cos^2x)$

$color(white)(RHS)=LHS$