Question

tan A tan A
+
sec A-1 sec A +1
= 2 cosec A.​

Answer 1

Correct Question :-

Prove :-

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

Solution :-

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

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

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

        $= \sf \dfrac{tanA \times 2secA}{tan^2A+1-1} \\\\= \dfrac{tanA \times 2secA}{tan^2A}\\\\= \dfrac{2secA}{tanA}$

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

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

$\bullet \bf \ cosecA = \dfrac{1}{sinA}$

         $= \sf 2cosecA\\\\= R.H.S$

Hence proved.