Question

prove that root 1+ sinA/ 1- sinA= sec A + tanA

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Answer 1

LHS = √(1 + sin∅)/√(1 - sin∅)

= √(1 + sin∅) × √(1 + sin∅)/√(1 -sin∅)×√(1 +sin∅)

= √(1 + sin∅)²/√(1 -sin²∅)

= (1 + sin∅)/√cos²∅

= (1 + sin∅)/cos∅

= 1/cos∅ + sin∅/cos∅

= sec∅ + tan∅ = RHS

hope it helps dear...

so hard...

Answer 2

To Prove :-

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

Solution :-

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

Multiply the numerator and denominator by $\sf 1+ sinA$,

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

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

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

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

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

Hence proved.