Question

prove that : sec theta +tan theta over sec theta - tan theta is equal to 1+sin theta over cos theta whole square

Answer 1

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Answer 2

Answer:

The proof is explained step-wise below :

Step-by-step explanation:

$\text{To Prove : }\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}=(\frac{1+\sin\theta}{\cos\theta})^2$

Taking L.H.S.

$\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}\\\\\text{On rationalizing the above fraction}\\\\\implies \frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}\times\frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta}\\\\\implies \frac{(\sec\theta+\tan\theta)^2}{1}\\\\\implies(\frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta})^2\\\\\implies(\frac{1+\sin\theta}{\cos\theta})^2$

= R.H.S

Hence Proved