Question

TO PROVE
[cosec theta-sin theta]*[sec theta-cos theta]*[tan theta+cot theta]=1

Answer 1

To Prove :-

$\sf (cosec \theta - sin \theta )( sec \theta - cos \theta )( tan \theta + cot \theta ) = 1$

Solution :-

$\sf L.H.S = ( cosec \theta - sin \theta )(sec \theta - cos \theta )( tan \theta + cot \theta )$

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

       $= \sf \left( \dfrac{1}{sin \theta}-sin \theta \right) \left( \dfrac{1}{cos \theta} - cos \theta } \right) \left( tan \theta + \dfrac{1}{tan \theta} \right) \\\\= \left( \dfrac{1- sin^2 \theta}{sin \theta} \right) \left( \dfrac{1-cos^2\theta}{cos\theta}\right) \left( \dfrac{tan^2\theta+1}{tan \theta} \right)$

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

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

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

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

Hence proved.