Question

√secA+tanA/secA-tanA=1+sinA/cosA​

Answer 1

Answer:

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Answer 2

To prove :

$\sqrt{\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}}= \frac{1+\sin\theta}{\cos\theta}$

Explanation:

$\sqrt{\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}}= \frac{1+\sin\theta}{\cos\theta}\\\\\text{squaring both sides we get}\\\\\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}= \left ( \frac{1+\sin\theta}{\cos\theta} \right )^2$

taking LHS

$\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}$

rationalizing we get

$\frac{\sec\theta+\tan\theta}{\sec\theta-\tan\theta}\times \frac{\sec\theta+\tan\theta}{\sec\theta+\tan\theta}\\\\=\frac{(\sec\theta+\tan\theta)^2}{1}\\\\=\left ( \frac{1}{\cos\theta}+\frac{\sin\theta}{\cos\theta} \right )\\\\=\left ( \frac{1+\sin\theta}{\cos\theta} \right )^2$

= RHS

Identity used : $\sec^2\theta - \tan^\theta = 1$

hence proved

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