Question

If $\begin{array}{c}\cot \theta +\cos \theta =a;\cot \theta -\cos \theta =b\end{array}$ then prove that $a^2-b^2=4\sqrt{ab}$

Answer 1

$a^2-b^2=(a-b)(a+b)=\left(2\cos \theta \right)\left(2\cot \theta \right)=4\cos \theta \cot \theta$

$4\sqrt{ab}=4\sqrt{({\cot }^2\theta -{\cos }^2\theta )}=4\sqrt{\frac{{\cos }^2\theta }{{\sin }^2\theta }-{\cos }^2\theta }=4\cos \theta \sqrt{\frac{1-{\sin }^2\theta }{{\sin }^2\theta }}=4\cos \theta \cot \theta$

Hence proved