Question

If $A$ and $B$ are satisfying the equation $3\sin x+4\cos x=1$, then $\cos A\cos B$ is

• A. $-\frac{9}{25}$
• B. $-\frac{8}{25}$
• C. $-\frac{15}{25}$
• D. $-\frac{16}{25}$

Answer 1

The correct option is B $-\frac{8}{25}$

$4\cos x+3\sin x=1$

$\Rightarrow 3\sin x=1-4\cos x$

Squaring the both sides we have

$\Rightarrow {\left(3\sin x\right)}^2={\left(1-4\cos x\right)}^2$

$9{\sin }^{2}x=1+16{\cos }^{2}x-8\cos x$

$\Rightarrow 9{\left(1-\cos x\right)}^2=1+16{\cos }^{2}x-8\cos x$

$\Rightarrow 9-9{\cos }^{2}x=1+16{\cos }^{2}x-8\cos x$

$\Rightarrow -9{\cos }^{2}x-16{\cos }^{2}x+8\cos x-1+9=0$

$\Rightarrow -25{\cos }^{2}x+8\cos x+8=0$

$\Rightarrow 25{\cos }^{2}x-8\cos x-8=0$

$\because$ $A$ and $B$ are satisfied

$\therefore \cos A.\cos B=\frac{c}{a}=\frac{-8}{25}$

$\therefore \cos A.\cos B=\frac{-8}{25}$