Question

If $(a^2-b^2)\sin \theta +2ab\cos \theta =a^2+b^2$ then find the value of $\tan \theta$

Answer 1

$\begin{array}{c}\left(a^2-b^2\right)\sin \theta +2ab\cos \theta =a^2+b^2\\ Leta=k\cos \varphi ,b=k\sin \varphi ,a^2+b^2=k^2,\tan \varphi =b/a\\ k^2\left({\cos }^2\varphi -si{n}^2\varphi \right)\sin \theta +2k^2\cos \varphi \sin \theta =k^2\\ \cos 2\varphi \sin \varphi +\sin 2\varphi \cos \theta =1\\ \sin \left(2\varphi +\theta \right)=1\\ \theta =\frac{\pi }{2}-2\varphi \\ \tan \theta =\cot 2\varphi \\ =\frac{1+{\tan }^2\varphi }{2\tan \varphi }=\frac{a^2+b^2}{2ab}\end{array}$