Question

$\begin{array}{c}\cos x+\cos y=\frac{4}{5},\cos x-\cos y=\frac{2}{7}\\ \Rightarrow 14\tan \left(\frac{x-y}{2}\right)+5\cot \left(\frac{x+y}{2}\right)=\end{array}$

• A. 0
• B. 1/4
• C. 5/4
• D. 3/4

Answer 1

The correct option is A 0

We have,

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

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

We know that

$\cos C+\cos D=2\cos \left(\frac{C+D}{2}\right)\cdot \cos \left(\frac{C-D}{2}\right)$

$\cos C-\cos D=2\sin \left(\frac{C+D}{2}\right)\cdot \sin \left(\frac{D-C}{2}\right)$

Therefore,

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

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

Now,

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

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

From equation (1) and (2), we get

$\frac{\cos \left(\frac{x+y}{2}\right)\cdot \cos \left(\frac{x-y}{2}\right)}{\sin \left(\frac{x+y}{2}\right)\cdot \sin \left(\frac{y-x}{2}\right)}=\frac{\frac{2}{5}}{\frac{1}{7}}$

$\frac{\cos \left(\frac{x+y}{2}\right)\cdot \cos \left(\frac{y-x}{2}\right)}{\sin \left(\frac{x+y}{2}\right)\cdot \sin \left(\frac{y-x}{2}\right)}=\frac{\frac{2}{5}}{\frac{1}{7}}$

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

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

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

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

Hence, the value is $0$.