Question

Sec theta +tan theta =p then show that sin theta =p2_1/p2+1

Answer 1

Given :-

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

To Prove :-

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

Solution :-

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

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

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

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

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

             $= \sf \dfrac{\dfrac{sin \theta}{cos\theta}}{ cos \theta} \\\\= \dfrac{sin \theta}{cos\theta} \times cos \theta \\\\= sin \theta\\\\= L.H.S$

L.H.S = R.H.S

Hence proved.