Question

Prove that $\sqrt{{\cot }^2\theta -{\cos }^2\theta }=\cot \theta \cos \theta$

Answer 1

$\sqrt{{\cot }^2\theta -{\cos }^2\theta }$

$=\sqrt{\frac{{\cos }^2\theta }{{\sin }^2\theta }-{\cos }^2\theta }$

$=\sqrt{{\cos }^2\theta (\frac{1}{{\sin }^2\theta }-1)}$

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

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

$=\cos \theta \times \left(\frac{\cos \theta }{\sin \theta }\right)$

$=\cos \theta \cot \theta$

Hence proved.