How do you verify the identity $\frac{sin θ}{1 - cot θ} + \frac{cos θ}{1 - tan θ} = sin θ + cos θ$ $sintheta/(1-cottheta)+costheta/(1-tantheta)=sintheta+costheta$ ?

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How do you verify the identity $\frac{sin θ}{1 - cot θ} + \frac{cos θ}{1 - tan θ} = sin θ + cos θ$ $sintheta/(1-cottheta)+costheta/(1-tantheta)=sintheta+costheta$ ?

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Answer 1 · 2016-09-06T13:57:34

LHS:

$= \frac{sin θ}{1 - \frac{cos θ}{sin θ}} + \frac{cos θ}{1 - \frac{sin θ}{cos θ}}$ $=sintheta/(1 - costheta/sintheta) + costheta/(1 - sintheta/costheta)$

$= \frac{sin θ}{\frac{sin θ - cos θ}{sin θ}} + \frac{cos θ}{\frac{cos θ - sin θ}{cos θ}}$ $=sintheta/((sin theta - costheta)/sintheta) + costheta/((costheta - sin theta)/costheta)$

$= \frac{sin θ sin θ}{sin θ - cos θ} + \frac{cos θ cos θ}{cos θ - sin θ}$ $=(sinthetasintheta)/(sin theta - costheta) + (costhetacostheta)/(costheta - sintheta)$

$= \frac{- {sin}^{2} θ + {cos}^{2} θ}{cos θ - sin θ}$ $=(-sin^2theta + cos^2theta)/(costheta - sin theta)$

$= \frac{{cos}^{2} θ - {sin}^{2} θ}{cos θ - sin θ}$ $=(cos^2theta- sin^2theta)/(costheta - sintheta$

$= \frac{(cos θ + sin θ) (cos θ - sin θ)}{cos θ - sin θ}$ $=((costheta + sin theta)(costheta - sin theta))/(costheta - sin theta)$

$=((costheta + sin theta)cancel(costheta - sin theta))/cancel(costheta - sin theta)$

$=costheta + sintheta$

$LHS = RHS$

Hopefully this helps!

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How do you verify the identity $\frac{sin θ}{1 - cot θ} + \frac{cos θ}{1 - tan θ} = sin θ + cos θ$ $sintheta/(1-cottheta)+costheta/(1-tantheta)=sintheta+costheta$ ?

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