Question

Prove that: 1+sinA/1-sinA= (secA+tanA)​ 2

Answer 1

TO PROVE :-

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

SOLUTION :-

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

Multiply numerator and denominator by $\sf 1+ sinA$ .

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

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

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

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

        $= \sf (secA+tanA)^2 \\\\= R.H.S$

Hence proved .