Question

tanA/1+secA - tanA/1-secA = ?

pls anwer this question quickly​

Answer 1

Step-by-step explanation:

i think the above photo wld help u in solving the qt

Answer 2

$\frac{tanA}{1 + secA} - \frac{tanA}{1 - secA} \\ \\ = tanA( \frac{1}{1 + secA } - \frac{1}{1 - secA} ) \\ \\ = tanA( \frac{1 - secA - 1 - secA}{1 - {sec}^{2} A} )\\ \\ = tanA( \frac{ - 2secA}{ - {tan}^{2}A } ) \\ \\ = 2 \frac{secA}{tanA} \\ \\ = 2 \: \frac{1}{cosA} . \frac{cosA}{sinA} \\ \\ = 2cosecA$