Question #1f5ff

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Question #1f5ff

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Answer 1 · 2017-03-12T22:07:20

Let us start with the LHS:

$L H S = (\frac{sin θ + cos θ}{sin θ}) - (\frac{cos θ - sin θ}{cos θ})$ $LHS = ((sintheta+costheta)/sintheta)-((costheta-sintheta)/costheta)$

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

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

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

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

$= \frac{1}{sin θ cos θ}$ $" " = (1)/(sinthetacostheta)$

$= \frac{1}{sin θ} \cdot \frac{1}{cos θ}$ $" " = 1/sintheta*1/costheta$

$= csc θ \cdot sec θ$ $" " = csctheta*sectheta$

$" " = RHS \ \ \ \$ QED

Answer 2 · 2017-03-13T11:50:05

$LHS=(sintheta+costheta)/sintheta-(costheta-sintheta)/costheta$

$=sintheta/sintheta+costheta/sintheta-costheta/costheta+sintheta/costheta$

$=cancel1+costheta/sintheta-cancel1+sintheta/costheta$

$=(cos^2theta+sin^2theta)/(sinthetacostheta)$

$=1/(sinthetacostheta)$

$=secthetacsctheta=RHS$

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Question #1f5ff

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