Question #8df79

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Answer 1 · 2017-01-31T17:04:49

$L H S = \frac{1 + sec θ}{sec θ}$ $LHS=(1+sectheta)/sectheta$

$= \frac{cos θ (1 + sec θ)}{cos θ sec θ}$ $=(costheta(1+sectheta))/(costhetasectheta)$

$= (cos θ + 1)$ $=(costheta+1)$

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

$= \frac{1 - {cos}^{2} θ}{1 - cos θ}$ $=(1-cos^2theta)/(1-costheta)$

$= \frac{{sin}^{2} θ}{1 - cos θ} = R H S$ $=sin^2theta/(1-costheta)=RHS$