Question

prove that tanA/1+secA-tanA/1-secA=2cosecA

Answer 1

Answer:

Step-by-step explanbioation:

Answer 2

Answer:

$\frac{tanA}{1+secA}-\frac{tanA}{1-secA}=2cosecA$

Step-by-step explanation:

$LHS = \frac{tanA}{1+secA}-\frac{tanA}{1-secA}$

$=tanA\big(\frac{1}{1+secA}-\frac{1}{1-secA}\big)$

$=tanA\big(\frac{(1-secA)-(1+secA)}{(1+secA)(1-secA)}\big)$

$=tanA\big(\frac{(1-secA-1-secA)}{(1^{2}-sec^{2}A)}\big)$

$=tanA\big(\frac{(-2secA)}{-(sec^{2}A-1)}\big)$

$=tanA\big(\frac{(-2secA)}{-tan^{2}A}\big)$

/* By Trigonometric identity:

sec²A-1 = tan²A */

$=\big(\frac{(2secA)}{tanA}\big)$

$=\frac{\frac{2}{cosA}}{\frac{sinA}{cosA}}$

$=\frac{2}{sinA}$

$=2cosecA$

$=RHS$

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