Question

$\cos x+\cos y=\frac{4}{5}$, $\cos x-\cos y=\frac{2}{7}$. The value of $14\tan \left(\frac{x-y}{2}\right)+5\cot \left(\frac{x+y}{2}\right)$ is

• A. $0$
• B. $\frac{1}{4}$
• C. $\frac{5}{4}$
• D. $\frac{3}{4}$

Answer 1

The correct option is A $0$

$\cos x+\cos y=\frac{4}{5}$

$2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{4}{5}$

$\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)=\frac{2}{5}.........\left(1\right)$

$\cos x-\cos y=\frac{2}{7}$

$-2\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)=\frac{2}{7}$

$-\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)=\frac{1}{7}.........\left(2\right)$

Dividing $\left(2\right)$ by $\left(1\right)$, we get

$-\frac{\sin \left(\frac{x+y}{2}\right)\sin \left(\frac{x-y}{2}\right)}{\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)}=\frac{5}{14}$

$\Rightarrow 14\tan \left(\frac{x+y}{2}\right)=-5\cot \left(\frac{x-y}{2}\right)$

$\Rightarrow 14\tan \left(\frac{x+y}{2}\right)+5\cot \left(\frac{x-y}{2}\right)=0$