Question

If in an obtuse angled triangle the obtuse angle is $\frac{3\pi }{4}$ and the other two angles are equal to two values of $\theta$ satisfying $a\tan \theta +b\sec \theta =c,$ when $\left|b\right|\le \sqrt{(a^2+c^2)}$, then $a^2-c^2$ is equal to

• A. $ac$
• B. $2ac$
• C. $\frac{a}{c}$
• D. none of these.

Answer 1

Answer: (B)

$2ac$

We know that $A+B+C=\pi \Rightarrow A+C=\frac{\pi }{4}$
$\Rightarrow \tan (A+C)=\tan \frac{\pi }{4}=1$
$\Rightarrow \frac{\tan A+\tan C}{1-\tan A\tan C}=1$
$\Rightarrow \tan A+\tan C=1-\tan A\tan C$ ............$\left(1\right)$
$\because a\tan \theta +b\sec \theta =c$
$\Rightarrow a\tan \theta -c=-b\sec \theta$
Squaring both sides, we get
$\Rightarrow {(a\tan \theta -c)}^2={(-b\sec \theta )}^2$
$\Rightarrow a^2{\tan }^2\theta +c^2-2ac\tan \theta -b^2{\sec }^2\theta =0$
$\Rightarrow (a^2-b^2){\tan }^2\theta -2ac\tan \theta +(c^2-b^2)=0$
Let $\tan A$ and $\tan C$ be the roots of the equation,
$\therefore \tan A+\tan C=\frac{2ac}{a^2-b^2}$ and
$\tan A\tan C=\frac{c^2-b^2}{a^2-b^2}$
From eqn$\left(1\right)$
$\frac{2ac}{a^2-b^2}=1-\frac{c^2-b^2}{a^2-b^2}$
$\Rightarrow \frac{2ac}{a^2-b^2}=\frac{a^2-b^2-c^2+b^2}{a^2-b^2}$
$\Rightarrow 2ac=a^2-c^2$