Question

In a triangle one angle is $\frac{3\pi }{4}$ and other two angles are two values of $\theta$ satisfying $a\tan \theta +b\sec \theta =c$ where $\left|b\right|\le \sqrt{a^2+c^2}$. Then $a^2-c^2$ is equal to

• A. $ac$
• B. $2ac$
• C. $\frac{a}{c}$
• D. $\frac{2a}{c}$

Answer 1

Answer: (B)

$2ac$

We have, $a\tan \theta +b\sec \theta =c\Rightarrow c\cos \theta -a\sin \theta =b$

Let $\alpha$ and $\beta$ be other two angles of the triangle

$\Rightarrow c\cos \alpha -a\sin \alpha =c\cos \beta -a\sin \beta$

$\Rightarrow c(\cos \alpha -\cos \beta )=a(\sin \alpha -\sin \beta )$

$\Rightarrow 2c\sin \frac{\alpha +\beta }{2}\sin \frac{\beta -\alpha }{2}=2a\cos \frac{\alpha +\beta }{2}\sin \frac{\alpha -\beta }{2}$

$\Rightarrow \tan \frac{\alpha +\beta }{2}=-\frac{a}{c}$

$\therefore \tan (\alpha +\beta )=\frac{2(-\frac{a}{c})}{1-\frac{a^2}{c^2}}\Rightarrow a^2-c^2=2ac$