Question

Prove the identity
tanA/1+secA - tanA/1-secA=2cosecA

Answer 1

To Prove :-

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

Solution :-

$\sf L.H.S = \dfrac{tanA}{1+secA} -\dfrac{tanA}{1-secA}$

       $= \sf \dfrac{tan(1-secA)-tanA(1+secA)}{(1+secA)(1-secA)} \\\\= \dfrac{tanA-tanAsecA-tanA-tanAsecA}{1-sec^2A} \\\\= \dfrac{-2tanAsecA}{1-sec^2A}$

Multiply the numerator and denominator by -1.

       $= \sf \dfrac{2tanAsecA}{sec^2A-1}$

$\bullet \bf \ sec^2A-1 = tan^2A$

        $= \sf \dfrac{2tanAsecA}{tan^2A} \\\\= \dfrac{2secA}{tanA}$

$\bullet \bf \ secA = \dfrac{1}{cosA} \\\\\bullet \ tanA = \dfrac{sinA}{cosA}$

         $= \sf \dfrac{\dfrac{2}{cosA}}{\dfrac{sinA}{cosA}} \\\\= \dfrac{2}{cosA} \times \dfrac{cosA}{sinA} \\\\= 2 \times \dfrac{1}{sinA} \\\\= 2 cosecA$

Hence proved.

Answer 2

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