Question

Prove :

√{(sec θ – 1)/(sec θ + 1)}  =  cosec θ - cot θ

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Answer 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.