Question

prove: cos theta ÷ sec theta - tan theta = 1 + sin theta​

Answer 1

TO PROVE :-

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

SOLUTION :-

L.H.S = $\sf\dfrac{cos\theta}{sec\theta-tan\theta}$

We know that,

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

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

We know that,

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

             $= \sf \dfrac{1-sin^2\theta}{1-sin\theta} \\\\= \dfrac{(1-sin\theta) \times (1+sin \theta)} {1-sin\theta}\\\\= 1+sin\theta \\\\= R.H.S$

Hence proved.