Question

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

Answer 1

Step 1: Simplify the given equation.

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

Dividing by $a^2+b^2$ throughout the equation we get,

$\Rightarrow \frac{(a^2-b^2)\sin \theta }{a^2+b^2}+\frac{2ab\cos \theta }{a^2+b^2}=1...\left(i\right)$

Say $\frac{a^2-b^2}{a^2+b^2}=\cos \alpha$$...\left(ii\right)$

Then $\sin \alpha =\sqrt{1-{\cos }^2\alpha }$

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

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

$\Rightarrow$ $\sin \alpha =\frac{2ab}{a^2+b^2}$$...\left(iii\right)$

Step 2: Solve further to get the required answer.

Now, substitute value of $\cos \alpha$ and $\sin \alpha$ in the equation $\left(i\right)$, we get,

$\Rightarrow$$\cos \alpha \sin \theta +\sin \alpha \cos \theta =1$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \sin \left(A+B\right)=\mathrm{sinAcosB}+\mathrm{cosAsinB}\right]$

$\Rightarrow$ $\sin (\alpha +\theta )=1$

$\Rightarrow$ $\sin (\alpha +\theta )=\sin \frac{\mathrm{\pi }}{2}$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \sin \frac{\pi }{2}=1\right]$

$\Rightarrow$ $\alpha +\theta =\frac{\mathrm{\pi }}{2}$

$\Rightarrow$ $\theta =\frac{\mathrm{\pi }}{2}-\alpha$

$\Rightarrow$ $\sin \theta =\sin \left(\frac{\mathrm{\pi }}{2}-\alpha \right)$

$\Rightarrow$ $\sin \theta =\cos \alpha$ $\mathbf{.}\mathbf{.}\mathbf{.}\left[\because \sin \left(\frac{\pi }{2}-A\right)=\mathrm{cosA}\right]$

$\Rightarrow$ $\sin \theta =\frac{a^2-b^2}{a^2+b^2}$ …[From (ii)]

Hence, the required answer is $\sin \theta =\frac{a^2-b^2}{a^2+b^2}$