Question

TAN THETA + SIN THETA/TAN THETA - SIN THETA =SEC THETA + 1 / SEC THETA - 1​

Answer 1

Answer:

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Answer 2

To Prove :-

$\sf \dfrac{tan \theta + sin \theta } {tan \theta - sin \theta}= \dfrac{sec \theta+1}{sec \theta -1 }$

Solution :-

$\sf L.H.S = \dfrac{tan \theta + sin \theta}{tan \theta - sin \theta }$

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

        $= \sf \dfrac{\dfrac{sin \theta }{cos\theta}+sin \theta}{\dfrac{sin \theta }{cos\theta}- sin \theta} \\\\= \dfrac{\dfrac{sin \theta+sin \theta cos \theta}{cos \theta}}{\dfrac{sin \theta - sin \theta cos \theta}{cos\theta}} \\\\= \dfrac{sin \theta+sin \theta cos\theta}{cos \theta} \times \dfrac{cos\theta}{sin\theta - sin \theta cos \theta} \\\\= \dfrac{sin \theta + sin \theta cos \theta }{sin \theta - sin \thets cos \theta} \\\\= \dfrac{sin \theta (1+cos\theta)}{sin \theta (1-cos \theta}$

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

Hence proved.