Question

(cosA/1+sinA+1+sinA/cisA)(cosA/1-sinA-1-sinA/cosA)=4tanA secA​

Answer 1

To Prove :-

$\sf \left( \dfrac{cosA}{1+sinA}+ \dfrac{1+sinA}{cosA} \right) \left( \dfrac{cosA}{1-sinA} - \dfrac{1-sinA}{cosA} \right) = 4tanAsecA$

Solution :-

$\sf L.H.S =\left( \dfrac{cosA}{1+sinA}+ \dfrac{1+sinA}{cosA} \right) \left( \dfrac{cosA}{1-sinA} - \dfrac{1-sinA}{cosA} \right)$

        $= \sf\left( \dfrac{cos^2A+(1+sinA)^2}{(1+sinA)cosA} \right) \left( \dfrac{cos^2A-(1-sinA)^2}{(1-sinA)cosA} \right) \\\\= \left( \dfrac{cos^2A+sin^2A+1+2sinA}{(1+sinA)cosA} \right) \left( \dfrac{cos^2A-1-sin^2A+2sinA }{(1-sinA)cosA} \right)$

$\bullet \bf \ cos^2A+sin^2A = 1 \\\\\bullet \ cos^2A - 1 = -sin^2A$

       $= \sf \left( \dfrac{1+1+2sinA}{(1+sinA)cosA} \right) \left( \dfrac{-sin^2A+2sinA-sin^2A}{(1-sinA)cosA} \right) \\\\= \dfrac{(2+2sinA) ( -2sin^2A+2sinA)}{(1-sin^2A)cos^2A}\\\\= \dfrac{2(1+sinA) 2sinA(1-sinA)}{cos^2A \times cos^2A} \\\\= \dfrac{4sinA(1-sin^2A)}{cos^4A}$

       $= \sf \dfrac{4sinA \times cos^2A}{cos^4A} \\\\= \dfrac{4sinA}{cos^2A}\\\\= 4 \times \dfrac{sinA}{cosA} \times \dfrac{1}{cosA} \\\\= 4tanA secA \\\\= R.H.S$

Hence proved.

Answer 2

See the attachment given below...hope it helps you