Question

Prove that √sec²

Answer 1

Given:

Consider the given statement is

`\sqrt{\sec^2\theta+\text{cosec}^2\theta}=\tan \theta+\cot \theta`

To prove:

The given statement.

Solution:

We have,

`\sqrt{\sec^2\theta+\text{cosec}^2\theta}=\tan \theta+\cot \theta`

Taking LHS, we get

`LHS=\sqrt{\sec^2\theta+\text{cosec}^2\theta}`

`LHS=\sqrt{(1)/(\cos^2\theta)+(1)/(\sin^2\theta)}`

`LHS=\sqrt{(\sin^2\theta+\cos^2\theta)/(\sin^2\theta\cos^2\theta)}`

`LHS=\sqrt{(1)/(\sin^2\theta\cos^2\theta)}`
`[\because \sin^2\theta+\cos^2\theta=1]`

`LHS=(1)/(\sin\theta\cos\theta)`

`LHS=(\sin^2\theta+\cos^2\theta)/(\sin\theta\cos\theta)`
`[\because \sin^2\theta+\cos^2\theta=1]`

`LHS=(\sin^2\theta)/(\sin\theta\cos\theta)+(\cos^2\theta)/(\sin\theta\cos\theta)`

`LHS=(\sin\theta)/(\cos\theta)+(\cos\theta)/(\sin\theta)`

`LHS=\tan \theta+\cot \theta`

`LHS=RHS`

Hence proved.