How do you simplify $\frac{sin a + tan a}{1 + cos a}$ $(sina+tana)/(1+cosa)$ ?

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How do you simplify $\frac{sin a + tan a}{1 + cos a}$ $(sina+tana)/(1+cosa)$ ?

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Answer 1 · 2016-10-11T03:16:33

$\frac{sin (a) + tan (a)}{1 + cos (a)} = tan (a)$ $(sin(a)+tan(a))/(1+cos(a))=tan(a)$

Explanation:

$\frac{sin (a) + tan (a)}{1 + cos (a)} = \frac{(sin (a) + tan (a)) (1 - cos (a))}{(1 + cos (a)) (1 - cos (a))}$ $(sin(a)+tan(a))/(1+cos(a)) = ((sin(a)+tan(a))(1-cos(a)))/((1+cos(a))(1-cos(a)))$

$= \frac{tan (a) - sin (a) cos (a)}{1 - {cos}^{2} (a)}$ $=(tan(a)-sin(a)cos(a))/(1-cos^2(a))$

$= \frac{tan (a) - sin (a) cos (a)}{{sin}^{2} (a)}$ $=(tan(a)-sin(a)cos(a))/sin^2(a)$

$= \frac{cot (a) (tan (a) - sin (a) cos (a))}{cot (a) {sin}^{2} (a)}$ $=(cot(a)(tan(a)-sin(a)cos(a)))/(cot(a)sin^2(a))$

$= \frac{1 - {cos}^{2} (a)}{cos (a) sin (a)}$ $=(1-cos^2(a))/(cos(a)sin(a))$

$= \frac{{sin}^{2} (a)}{cos (a) sin (a)}$ $=sin^2(a)/(cos(a)sin(a))$

$= \frac{sin (a)}{cos (a)}$ $=sin(a)/cos(a)$

$= tan (a)$ $=tan(a)$

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