Question

prove that p^2-1÷p^2+1= sin tetha, if sec tetha + tan tetha =p​

Answer 1

Given :-

$\sf sec\theta+tan\theta = p$

To Prove :-

$\sf \dfrac{p^2-1}{p^2+1} = sin \theta$

Solution :-

$\sf L.H.S = \dfrac{p^2-1}{p^2+1}$

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

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

             $= \sf \dfrac{tan^2 \theta+tan^2 \theta+ 2sec \theta tan \theta }{sec^2 \theta + sec^2 \theta + 2sec \theta tan\theta } \\\\= \dfrac{2tan^2\theta+2sec\theta tan \theta }{ 2sec^2 \theta +2sec \theta tan \theta } \\\\= \dfrac{2tan\theta (tan \theta +sec \theta )}{2sec \theta ( sec \theta+tan \theta)} \\\\= \dfrac{tan\theta}{sec\theta} \\\\= \dfrac{\dfrac{sin \theta}{cos\theta}}{ \dfrac{1}{cos\theta }} \\\\= \dfrac{sin \theta}{cos \theta} \times cos \theta \\\\= sin \theta \\\\= R.H.S$

Hence proved.