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 $a\tan \theta +b\sec \theta =c$, where $|b|\le \sqrt{a^2+c^2}$, then find the value of $a^2-c^2$ is equal to

• A. $2ac$
• B. $2bc$
• C. $ab$
• D. $ac$

Answer 1

The correct option is A $2ac$

Given 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 $a\tan \theta +b\sec \theta =c$.

$\Rightarrow {\theta }_1$ and ${\theta }_2$ satisfies $a\tan \theta +b\sec \theta =c$ and ${\theta }_1+{\theta }_2=\frac{\pi }{4}$

$b\sec \theta =c-b\tan \theta$

squaring on both sides and simplifying gives

$(a^2-b^2){\tan }^2\theta -2ac\tan \theta -(b^2-c^2)=0$ -----(1)

As, ${\theta }_1+{\theta }_2=\frac{\pi }{4}$

$\tan {\theta }_1+\tan {\theta }_2=1-\tan {\theta }_1\tan {\theta }_2$

$\Rightarrow \frac{2ac}{a^2-b^2}=1+\frac{b^2-c^2}{a^2-b^2}$ from (1)

$\Rightarrow \frac{2ac}{a^2-b^2}=\frac{a^2-c^2}{a^2-b^2}$

$\therefore a^2-c^2=2ac$