Question

In a $\Delta ABC,\tan A$ and $\tan B$ are the roots of the equation $ab(x^2+1)=c^{2}x,$ where $a,b$ and $c$ are the sides of the triangle. Then

• A. $\tan (A-B)=-\frac{a^2+b^2}{2ab}$
• B. $\cot C=0$
• C. ${\sin }^{2}A+{\sin }^{2}B=1$
• D. none of these

Answer 1

The correct option is A $\tan (A-B)=-\frac{a^2+b^2}{2ab}$

$ab{x}^2-c^{2}x+ab=0$
Now sum of roots
$=\tan A+\tan B$
$=\frac{c^2}{ab}$
And product of roots
$=\tan A.\tan B$
$=1$.
Hence
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A.\tan B}$
$=\infty$
Hence
$A+B=\frac{\pi }{2}$.
Or
$\pi -C=\frac{\pi }{2}$
Or
$C=\frac{\pi }{2}$.
Hence the triangle is a right angled triangle.
Hence
$\tan A=\cot B$
Now
$\tan A+\tan B=\frac{c^2}{ab}$
Or
$\tan A+\cot A=\frac{c^2}{ab}$
Or
$\frac{{\tan }^2+1}{\tan A}=\frac{c^2}{ab}$
Or
$\frac{{\tan }^2+1}{2\tan A}=\frac{c^2}{2ab}$
Or
$\cot 2A=\frac{c^2}{2ab}$ ...(i)
Now
$\tan (A-B)$
$=\tan (A-(\frac{\pi }{2}-A\left)\right)$

$=\tan (2A-\frac{\pi }{2})$

$=-\cot 2A$

$=\frac{-c^2}{2ab}$
Now by pythagoras theorem, $c^2=a^2+b^2$
Or
$\tan (A-B)=-\frac{a^2+b^2}{2ab}$