Question

$If\left(a^2-b^2\right)\sin \theta +2ab\cos \theta =a^2+b^{2}find\tan \theta$

Answer 1

$\left(a^2-b^2\right)\sin \theta +2ab\cos \theta =a^2+b^2\left(dividingby\cos \theta \right)$

= $\left(a^2-b^2\right)\frac{\sin \theta }{\cos \theta }+2ab\frac{\cos \theta }{\cos \theta }=\frac{a^2+b^2}{\cos \theta }$

= $\left(a^2-b^2\right)\tan \theta +2ab=\left(a^2+b^2\right)\sec \theta$

= $squaringbothsideweget$

= ${\left(a^2-b^2\right)}^2{\tan }^2\theta +4a^2{b}^2+4ab\left(a^2-b^2\right)\tan \theta ={\left(a^2+b^2\right)}^2{\sec }^2\theta$

= ${\left(a^2-b^2\right)}^2{\tan }^2\theta +4a^2{b}^2+4ab\left(a^2-b^2\right)\tan \theta ={\left(a^2+b^2\right)}^2\left(1+{\tan }^2\theta \right)$

= $\left[{\left(a^2-b^2\right)}^2-{\left(a^2+b^2\right)}^2\right]{\tan }^2\theta +4a^2{b}^2+4ab\left(a^2-b^2\right)\tan \theta -{\left(a^2+b^2\right)}^2=0$

= $\left[\left(a^2-b^2+a^2+b^2\right)\left(a^2-b^2-a^2-b^2\right)\right]{\tan }^2\theta +4ab\left(a^2-b^2\right)\tan \theta +4a^2{b}^2-\left(a^4+b^4+2a^2{b}^2\right)=0$

$-4a^2{b}^2{\tan }^2\theta +4ab\left(a^2-b^2\right)\tan \theta +2a^2{b}^2-a^4-b^4=0$

$+{\tan }^2\theta -\left(\frac{a^2-b^2}{ab}\right)\tan \theta -\frac{1}{2}+\frac{a^2}{4b^2}+\frac{b^2}{4a^2}=0$

${\tan }^2\theta +\left(\frac{b^2-a^2}{ab}\right)\tan \theta -\frac{1}{2}+\frac{a^2}{4b^2}+\frac{b^2}{4a^2}=0$

$put\tan \theta =u,then$

$u^2+\left(\frac{b^2-a^2}{ab}\right)u-\frac{1}{2}+\frac{a^2}{4b^2}+\frac{b^2}{4a^2}=0$

$u=\frac{-\left(\frac{b^2-a^2}{ab}\right)\pm \sqrt{{\left(\frac{b^2-a^2}{ab}\right)}^2-4\left(\frac{a^2}{4b^2}+\frac{b^2}{4a^2}-\frac{1}{2}\right)}}{2}$

$=\frac{\frac{a^2-b^2}{ab}\pm \sqrt{\frac{4{\left(b^2-a^2\right)}^2-4\left(a^4+b^4\right)-2a^2{b}^2}{4a^2{b}^2}}}{2}$

$=\frac{\frac{a^2-b^2}{ab}\pm \sqrt{\frac{4b^4+4a^4-8a^2{b}^2-4a^4-4b^4+8a^2{b}^2}{4a^2{b}^2}}}{2}$

$\tan \theta =\frac{a^2-b^2}{2ab}$