Question

if $(a^2-b^2)\sin \theta +2ab\cos \theta =a^2+b^2$, then prove $\tan \theta =\frac{a^2-b^2}{2ab}$

Answer 1

According to question.......

$\begin{array}{c}(a^2-b^2)\sin \theta +2ab\cos \theta =a^2+b^2\\ \Rightarrow (a^2+b^2)-(a^2-b^2)\sin \theta =2ab\cos \theta \\ Now,squaringbothside,\\ \Rightarrow {\left((a^2+b^2)-(a^2-b^2)\sin \theta \right)}^2=(2ab\cos \theta {)}^2\\ \Rightarrow (a^2+b^2{)}^2+(a^2-b^2{)}^2{\sin }^2\theta -2(a^2+b^2)(a^2-b^2)\sin \theta =4a^2{b}^{{}^2}{\cos }^2\theta \\ \Rightarrow (a^2+b^2{)}^2+(a^2-b^2{)}^2{\sin }^2\theta -2(a^2+b^2)(a^2-b^2)\sin \theta -4a^2{b}^{{}^2}(1-si{n}^2\theta )=0\\ \Rightarrow (a^2+b^2{)}^2+(a^2-b^2{)}^2{\sin }^2\theta -2(a^2+b^2)(a^2-b^2)\sin \theta -4a^2{b}^2+4a^2{b}^2{\sin }^2\theta )=0\\ \Rightarrow \left[{(a^2-b^2)}^2+4a^2{b}^2\right]{\sin }^2\theta -2(a^2+b^2{)}^2(a^2-b^2)\sin \theta +\left[{(a^2+b^2)}^2-4a^2{b}^2\right]=0\\ \Rightarrow (a^2+b^2{)}^2{\sin }^2\theta -2(a^2+b^2{)}^2(a^2-b^2)\sin \theta +(a^2-b^2{)}^2=0\\ \Rightarrow {\left[{(a^2+b^2)}^2\sin \theta -(a^2-b^2)\right]}^2=0\\ \Rightarrow (a^2+b^2)\sin \theta -(a^2-b^2)=0\\ \Rightarrow \sin \theta =\frac{(a^2-b^2)}{(a^2+b^2)}\\ \Rightarrow \cos ec\theta =\frac{a^2+b^2}{a^2-b^2}\\ Now,\\ {\cot }^2\theta =\cos e{c}^2\theta -1\\ \Rightarrow {\cot }^2\theta ={\left(\frac{a^2+b^2}{a^2-b^2}\right)}^2-1\\ \Rightarrow {\cot }^2\theta =\frac{{(a^2+b^2)}^2-{(a^2-b^2)}^2}{{(a^2-b^2)}^2}=\frac{(a^2+b^2-a^2+b^2)(a^2+b^2+a^2-b^2)}{{(a^2-b^2)}^2}\\ \Rightarrow {\cot }^2\theta =\frac{\left(2a^2\right)\left(2b^2\right)}{{(a^2-b^2)}^2}=\frac{4a^2{b}^2}{{(a^2-b^2)}^2}\\ \Rightarrow \cot \theta =\sqrt{\frac{4a^2{b}^2}{{(a^2-b^2)}^2}}=\frac{2ab}{a^2+b^2}\\ \Rightarrow \tan \theta =\frac{a^2+b^2}{2ab}|\cot \theta =\frac{1}{\tan \theta }\\ sothatweprove:\tan \theta =\frac{a^2+b^2}{2ab}\end{array}$