Math, asked by kishu29, 4 months ago

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Answered by BrainlyEmpire
1

To Prove :-

\sf \sqrt{\dfrac{sec \theta -1}{sec\theta+1}} = cosec \theta - cot \theta

Solution :-

\sf L.H.S = \sqrt{\dfrac{sec\theta -1 }{sec\theta+1}}

Multiply denominator and numerator by \sf sec\theta - 1.

         = \sf \sqrt{\dfrac{(sec\theta - 1)(sec\theta-1)}{(sec\theta+1)(sec\theta-1)}} \\\\= \sqrt{\dfrac{(sec \theta-1)^2}{sec^2 \theta-1}}        

\bullet \bf \ sec^2 \theta - 1 = tan^2 \theta

          = \sf \sqrt{\dfrac{(sec\theta-1)^2}{tan^2 \theta }} \\\\= \sqrt{\left( \dfrac{sec\theta}{tan \theta} - \dfrac{1}{tan\theta} \right)^2}}

\bullet \bf \ sec \theta = \dfrac{1}{cos \theta } \\\\\bullet \ tan \theta = \dfrac{sin \theta }{cos \theta } \\\\\bullet \ cot \theta = \dfrac{1}{tan \theta }

\bullet \ \bf cosec \theta = \dfrac{1}{sin \theta}

            = \sf \sqrt{\left(\dfrac{\dfrac{1}{cos\theta}}{\dfrac{sin \theta }{cos \theta }}- cot \theta \right)^2 } \\\\= \sqrt{ \left(\dfrac{1}{ sin \theta } -cot\theta \right)^2 } \\\\= \sqrt{(cosec\theta - cot \theta)^2}

            = \sf cosec \theta - cot \theta \\\\= R.H.S

Hence proved.

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