Question

(1+cot theta - cosec theta) ( 1+ tan theta + sec theta) = 2​

Answer 1

To Prove :-

$\sf (1+cot \theta- cosec \theta)(1+tan \theta+sec \theta )= 2$

Solution :-

$\sf L.H.S = (1+cot \theta-cosec \theta)(1+tan \theta+sec \theta)$

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

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

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

$\bullet \bf \ sin^2 \theta +cos ^2 \theta$

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

Hence proved.