How do you prove $(1 + csc A) (1 - sin A) = cot A cos A$ $(1 + csc A)(1 - sin A) = cotA cosA$ ?

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Answer 1 · 2015-05-02T22:11:52

Remember that
$csc (A) = \frac{1}{sin (a)}$ $csc(A) = 1/sin(a)$
and
$cot (A) = \frac{cos (A)}{sin (A)}$ $cot(A) = cos(A)/sin(A)$

Therefore
$1 + csc (A)) (1 - sin (A))$ $1+csc(A))(1-sin(A))$

$= 1 + \frac{1}{sin (A)} - sin (A) - \frac{sin (A)}{sin (A)}$ $=1+1/sin(A)-sin(A)-sin(A)/sin(A)$

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

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

$= \frac{cos (A)}{sin (A)} \cdot cos (A)$ $= cos(A)/sin(A) * cos(A)$

$= cot (A) \cdot cos (A)$ $=cot(A)*cos(A)$

How do you prove (1+cscA)(1−sinA)=cotAcosA(1 + csc A)(1 - sin A) = cotA cosA?