Question

If $(a^2-b^2)\sin \theta +2ab\cos \theta =a^2+b^2$, find $\tan \theta$.

Answer 1

If, $(a^2-b^2)\sin \theta +2ab\cos \theta =a^2+b^2$

Let, $x=a^2-b^2$ and $y=a^2+b^2$

Then, $4a^2{b}^2=(a^2+b^2{)}^2-(a^2-b^2{)}^2=y^2-x^2$

$⟹x\sin \theta +2ab\cos \theta =y$

$⟹(x\sin \theta +2ab\cos \theta {)}^2=y^2$

$⟹x^2{\sin }^2\theta +4a^2{b}^2{\cos }^2\theta +4abx\sin \theta \cos \theta =y^2$

$⟹x^2{\sin }^2\theta +(y^2-x^2){\cos }^2\theta +4abx\sin \theta \cos \theta =y^2$

$⟹x^2{\sin }^{\theta }+y^2{\cos }^2\theta -x^2{\cos }^2\theta +4abx\sin \theta \cos \theta =y^2$

$⟹x^2{\sin }^2\theta +y^2{\cos }^2\theta -y^2-x^2{\cos }^2\theta +4abx\sin \theta \cos \theta =0$

$⟹x^2{\sin }^2\theta +y^2({\cos }^2\theta -1)-x^2{\cos }^2\theta +4abx\sin \theta \cos \theta =0$

$⟹x^2{\sin }^2\theta -y^2{\sin }^2\theta -x^2{\cos }^2\theta +4abx\sin \theta \cos \theta =0$

$⟹-{\sin }^2\theta (y^2-x^2)-x^2{\cos }^2\theta +4abx\sin \theta \cos \theta =0$

$⟹4a^2{b}^2{\sin }^2\theta +x^2{\cos }^2\theta -4abx\sin \theta \cos \theta =0$

$⟹(2ab\sin \theta -x\cos \theta {)}^2=0$

$⟹2ab\sin \theta =x\cos \theta$

$⟹\frac{\sin \theta }{\cos \theta }=\frac{x}{2ab}$

$⟹\tan \theta =\frac{a^2-b^2}{2ab}$

$⟹cosec\theta =\frac{\sqrt{{{(a}^2-b^2)}^2-4a^2{b}^2}}{a^2-b^2}$

$=\frac{\sqrt{a^4-2a^2{b}^2+b^4+4a^2{b}^2}}{a^2-b^2}$

$⟹\frac{\sqrt{a^4+2a^2{b}^2+b^4}}{a^2-b^2}=\frac{(a^2+b^2)}{(a^2-b^2)}$