Question

Prove that : $\sqrt{(se{c}^2\Theta +cose{c}^2}=tan\Theta +cot\Theta .$

Answer 1

$\sqrt{({\sec }^2\theta +{\csc }^2\theta )}=\tan \theta +\cot \theta$

$L.H.S\Rightarrow \sqrt{\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }}=\sqrt{\frac{{\sin }^2\theta +{\cos }^2\theta }{{\sin }^2\theta .{\cos }^2\theta }}=\sqrt{\frac{1}{{\sin }^2\theta .{\cos }^2\theta }}$

$=\frac{1}{{\sin }^2\theta .\cos \theta }-----LHS$

$R.H.S:\tan \theta +\cot \theta =\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }=\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta .\cos \theta }$

$=\frac{1}{\sin \theta .\cos \theta }------R.H.S$

$\therefore LHS=RHS$