Respuesta :
Given:
Consider the given statement is
[tex]\sqrt{\sec^2\theta+\text{cosec}^2\theta}=\tan \theta+\cot \theta[/tex]
To prove:
The given statement.
Solution:
We have,
[tex]\sqrt{\sec^2\theta+\text{cosec}^2\theta}=\tan \theta+\cot \theta[/tex]
Taking LHS, we get
[tex]LHS=\sqrt{\sec^2\theta+\text{cosec}^2\theta}[/tex]
[tex]LHS=\sqrt{\dfrac{1}{\cos^2\theta}+\dfrac{1}{\sin^2\theta}}[/tex]
[tex]LHS=\sqrt{\dfrac{\sin^2\theta+\cos^2\theta}{\sin^2\theta\cos^2\theta}}[/tex]
[tex]LHS=\sqrt{\dfrac{1}{\sin^2\theta\cos^2\theta}}[/tex] [tex][\because \sin^2\theta+\cos^2\theta=1][/tex]
[tex]LHS=\dfrac{1}{\sin\theta\cos\theta}[/tex]
[tex]LHS=\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}[/tex] [tex][\because \sin^2\theta+\cos^2\theta=1][/tex]
[tex]LHS=\dfrac{\sin^2\theta}{\sin\theta\cos\theta}+\dfrac{\cos^2\theta}{\sin\theta\cos\theta}[/tex]
[tex]LHS=\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}[/tex]
[tex]LHS=\tan \theta+\cot \theta[/tex]
[tex]LHS=RHS[/tex]
Hence proved.
