Question

If Sec theta +tan theta= n prove that sin theta= n^2-1/n^2+1

Answer 1

Given :-

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

To Prove :-

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

Solution :-

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

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

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

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

         $= \sf sin \theta \\\\= L.H.S$

R.H.S = L.H.S

Hence proved.