Question

Prove the following trigonometric identities:
cosAcosecA-sinAsecA/cosA+sinA=cosecA-secA

Answer 1

$\text{Consider,}$

$\displaystyle\frac{cosA\;cosecA-sinA\;secA}{cosA+sinA}$

$=\displaystyle\frac{cosA(\frac{1}{sinA})-sinA(\frac{1}{cosA})}{cosA+sinA}$

$=\displaystyle\frac{\frac{cosA}{sinA}-\frac{sinA}{cosA}}{cosA+sinA}$

$=\displaystyle\frac{\frac{cos^2A-sin^2A}{sinA\;cosA}}{cosA+sinA}$

$=\displaystyle\frac{\frac{(cosA-sinA)(cosA+sinA)}{sinA\;cosA}}{cosA+sinA}$

$=\displaystyle\frac{cosA-sinA}{sinA\;cosA}$

$=\displaystyle\frac{cosA}{sinA\;cosA}-\frac{sinA}{sinA\;cosA}$

$=\displaystyle\frac{1}{sinA}-\frac{1}{cosA}$

$=\displaystyle\;cosecA-secA$

$\therefore\bf\displaystyle\frac{cosA\;cosecA-sinA\;secA}{cosA+sinA}=cosecA-secA$

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