Question

If $a^2(1-\sin \theta )+b^2(1+\sin \theta )=2ab\cos \theta$, then the value of $\tan \theta$ is

• A. $\frac{a^2-b^2}{a^2+b^2}$
• B. $2ab$
• C. $\frac{b^2-a^2}{2ab}$
• D. $\frac{a^2-b^2}{2ab}$

Answer 1

The correct option is D $\frac{a^2-b^2}{2ab}$

$a^2(1-\sin \theta )+b^2(1+\sin \theta )=2ab\cos \theta$
$\Rightarrow a^2(\sec \theta -\tan \theta )+b^2(\sec \theta +\tan \theta )=2ab$
$\Rightarrow \frac{a}{b}(\sec \theta -\tan \theta )+\frac{b}{a}\frac{1}{(\sec \theta -\tan \theta )}=2$
Put $\frac{a}{b}(\sec \theta -\tan \theta )=t$
$\Rightarrow t+\frac{1}{t}=2$
$\Rightarrow t^2+1=2t$
$\Rightarrow t^2-2t+1=0$
$\Rightarrow t=1$

Therefore, $\frac{a}{b}(\sec \theta -\tan \theta )=1$
$\Rightarrow \sec \theta -\tan \theta =\frac{b}{a}...\left(1\right)$
We know that, ${\sec }^2\theta -{\tan }^2\theta =1$
$\Rightarrow (\sec \theta +\tan \theta )(\sec \theta -\tan \theta )=1$
$\Rightarrow \sec \theta +\tan \theta =\frac{a}{b}...\left(2\right)$
From equation $\left(1\right)$ and $\left(2\right)$,
$\tan \theta =\frac{a^2-b^2}{2ab}$