Question

State the whether given statement is true or false

Prove that:$\sqrt{{\sec }^2\theta +cose{c}^2\theta }=\tan \theta +\cot \theta$.

• A. True
• B. False

Answer 1

The correct option is A True

$\sqrt{{\sec }^2\theta +cose{c}^2\theta }=\tan \theta +\cot \theta$

$\Rightarrow$ By L.H.S

$\sqrt{{\sec }^2\theta +cose{c}^2\theta }=\sqrt{\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }}=\sqrt{\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta {\cos }^2\theta }}$

$=\sqrt{\frac{1}{{\sin }^2\theta {\cos }^2\theta }}=\frac{1}{\sin \theta \cdot \cos \theta }$ ……$\left(1\right)$

By R.H.S.

$\tan \theta +\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }$

$=\frac{1}{\sin \theta \cos \theta }$ ………….$\left(2\right)$

So: L.H.S $=$ R.H.S.