Question

question in the attachment.solve it​

Answer 1

To Prove :-

$\sf \sqrt{\dfrac{1-sinA}{1+sinA}} = secA-tanA$

Solution :-

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

Multiply numerator and denominator by $\sf 1 -sinA$,

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

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

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

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

        $= \sf \sqrt{(secA-tanA)^2}\\\\= secA- tanA\\\\= R.H.S$

Hence proved!

Answer 2

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refer the attachment :)