Question

Prove $\sqrt{\frac{1+{\sin }^2\theta \cdot {\sec }^2\theta }{1+{\cos }^2\theta \cdot {\text{cosec}}^2\theta }}=\tan \theta$.

Answer 1

$\sqrt{\frac{1+{\sin }^2\theta \cdot {\sec }^2\theta }{1+{\cos }^2\theta {\csc }^2\theta }}=\tan \theta \sqrt{\frac{1+\frac{{\sin }^2\theta }{{\cos }^2\theta }}{1+\frac{{\cos }^2\theta }{{\sin }^2\theta }}}\sqrt{\frac{1+{\tan }^2\theta }{1+{\cot }^2\theta }}\sqrt{\frac{{\sec }^2\theta }{{\csc }^2\theta }}\sqrt{\frac{{\sin }^2\theta }{{\cos }^2\theta }}\sqrt{{\tan }^2\theta }=\tan \theta$