Question

If $0<A+B<\frac{\pi }{2}$ and $\tan A,\tan B$ are the roots of the equation $3x^2-12x-6=0,$ then the numerical value of $\sin (A+B)\cos (A+B)-\sec (A+B)\text{cosec}(A+B)$ is

• A. $-\frac{54}{125}$
• B. $\frac{481}{300}$
• C. $\frac{54}{125}$
• D. $-\frac{481}{300}$

Answer 1

The correct option is D $-\frac{481}{300}$

$3x^2-12x-6=0$
Now, the sum and product of the roots is,
$\tan A+\tan B=4,\tan A\tan B=-2$
$\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}\Rightarrow \tan (A+B)=\frac{4}{1+2}=\frac{4}{3}$
So, $\sin (A+B)=\frac{4}{5},\cos (A+B)=\frac{3}{5}$
Now, $\sin (A+B)\cos (A+B)-\sec (A+B)\text{cosec}(A+B)$
$=\sin (A+B)\cos (A+B)-\frac{1}{\sin (A+B)\cos (A+B)}$
$=\frac{4}{5}\times \frac{3}{5}-\frac{5}{4}\times \frac{5}{3}=-\frac{481}{300}$