Question

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

Answer 1

Answer: (A), (B), (C)

$\cot C=0$
${\sin }^{2}A+{\sin }^{2}B=1$
$\tan (A-B)=\frac{a^2-b^2}{2ab}$

$ab{x}^2-c^{2}x+ab=0$
$\tan A+\tan B=\frac{+c^2}{ab}-\left(1\right)$
$\tan A\tan B=1-\left(2\right)$ $\Rightarrow C={90}^o$

$DeltaABC$ is rightangle triangle.

$\to c^2=a^2+b^2$
$tan(A-B)=\frac{(\tan A-\tan B)}{(1+\tan A\tan B)}$

$A+B+C=\pi$
$A+B=\frac{\pi }{2}$...........(From eq.(2))
$\tan (A-B)=\frac{\sqrt{(\tan A+\tan B{)}^2-4\tan A\tan B}}{1+1}$

$=\frac{\sqrt{(\frac{c^2}{a^2{b}^2}{)}^2-4}}{2}$
$=\frac{\sqrt{\frac{(a^2+b^2{)}^2}{a^2{b}^2}-4}}{2}$
$=\frac{a^2-b^2}{2ab}$

Also $\cot C=\cot 90=1$

$\sin A=\frac{a}{c},\sin B=\frac{b}{c}$

$si{n}^{2}A+si{n}^{2}B=\frac{a^2+b^2}{c^2}=1$