Question

prove that : root of 1+sin2theta sec2theta/1+cos2theta cosec2theta=tantheta​

Answer 1

To Prove :-

$\sf \sqrt{\dfrac{1+sin^2 \theta sec^2 \theta }{1+cos^2 \theta cosec^2 \theta}} = tan \theta$

Solution :-

$\sf L.H.S = \sqrt{\dfrac{1+sin^2 \theta sec^2 \theta }{1+cos^2 \theta cosec ^2\theta }}$

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

       $= \sf \sqrt{\dfrac{1+ sin^2 \theta \times \dfrac{1}{cos^2 \theta }}{1+cos^2 \theta \times \dfrac{1}{sin^2 \theta }}} \\\\= \sqrt{\dfrac{1+ \dfrac{sin^2 \theta }{cos^2 \theta }}{1+ \dfrac{cos^2 \theta }{sin^2 \theta }}}$

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

        $= \sf \sqrt{\dfrac{1+tan^2 \theta }{1+cot^2 \theta }}$

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

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

Hence proved.