How do you verify the identity $(costhetacottheta)/(1-sintheta)-1=csctheta$ ?

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Answer 1 · 2017-01-17T07:33:52

See proof below

We need

$cottheta=costheta/sintheta$

$csctheta=1/sintheta$

$cos^2theta+sin^2theta=1$

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

Therefore,

$LHS=(costhetacottheta)/(1-sintheta)-1$

$=(costheta*costheta/sintheta)/(1-sintheta)-1$

$=cos^2theta/(sintheta(1-sintheta))-1$

$=(1-sin^2theta)/(sintheta(1-sintheta))-1$

$=((1+sintheta)cancel(1-sintheta))/(sinthetacancel(1-sintheta))-1$

$=(1+sintheta)/sintheta-1$

$=(1+sintheta-sintheta)/sintheta$

$=1/sintheta$

$=csctheta$

$=RHS$

$QED$

How do you verify the identity (costhetacottheta)/(1-sintheta)-1=csctheta(costhetacottheta)/(1-sintheta)-1=csctheta?