Question

If $\tan \alpha ,\tan \beta$ satisfy (1) and $\cos \gamma ,\cos \delta$ satisfy (2) then $\tan \alpha \tan \beta +\cos \gamma +\cos \delta$ can be equal to

• A. $-1$
• B. $-\frac{5}{3}+\frac{2}{\sqrt{13}}$
• C. $\frac{5}{3}-\frac{2}{\sqrt{13}}$
• D. $-\frac{5}{3}-\frac{2}{\sqrt{13}}$

Answer 1

Answer: (B), (D)

The correct options are
B $-\frac{5}{3}+\frac{2}{\sqrt{13}}$
D $-\frac{5}{3}-\frac{2}{\sqrt{13}}$

$5{\sin }^{2}x+3\sin x\cos x-3{\cos }^{2}x=2\Rightarrow 3{\sin }^{2}x+2-2{\cos }^{2}x+3\sin x\cos x-3{\cos }^{2}x=2\Rightarrow 3{\sin }^{2}x+3inx\cos x-3{\cos }^{2}x=0\Rightarrow 3{\tan }^{2}x+3\tan x-5=0\Rightarrow \tan x=\frac{-3\pm \sqrt{69}}{6}$
$\Rightarrow \tan \alpha =\frac{-3+\sqrt{69}}{6},\tan \beta =\frac{-3-\sqrt{69}}{6}$ ..(1)

${\sin }^{2}x-\cos 2x=2-\sin 2x\Rightarrow 1-{\cos }^{2}x-2{\cos }^{2}x+1=2-2\sin x\cos x\Rightarrow 3{\cos }^{2}x-2\sin x\cos x=0\Rightarrow \cos x(3\cos x-2\sin x)=0\Rightarrow \cos x=0,\cos x=\pm \frac{2}{\sqrt{13}}$
$\Rightarrow \cos \gamma =0,\cos \delta =\pm \frac{2}{\sqrt{13}}$ ...(2)
From (1) and (2)
$\tan \alpha .\tan \beta +\cos \gamma +\cos \delta =\left(\frac{-3+\sqrt{69}}{6}\right)\left(\frac{-3-\sqrt{69}}{6}\right)+0\pm \frac{2}{\sqrt{13}}=-\frac{5}{3}\pm \frac{2}{\sqrt{13}}$