Question #fb6ad

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Answer 1 · 2016-05-07T20:12:11

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Right Side $=(sinx-cscx)/(cosx-cotx)$

$=(sinx-1/sinx)/(cosx-cosx/sinx)$

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

$=(sin^2x-1)/sinx xx sinx/(cosx sinx-cosx)$

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

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

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

$=cosx/(1-sinx)$

$=$ Left Side