Question

if cos x + cos y =4/5 and cos x-cos y =2/7. find the values of 14 tan x-y/2 + 5 cot x+y/2.

Answer 1

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Answer 2

Answer:

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

Step-by-step explanation:

Given : If $\cos x+\cos y=\frac{4}{5}$  and $\cos x-\cos y=\frac{2}{7}$  

To find : The value of $14\tan(\frac{x-y}{2})+5\cot\frac{x+y}{2}$?

Solution :

If $\cos x+\cos y=\frac{4}{5}$  and $\cos x-\cos y=\frac{2}{7}$  

Applying trigonometric identities,

$\cos x+\cos y=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$

$\frac{4}{5}=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ .....(1)

$\cos x-\cos y=-2\sin(\frac{x-y}{2})\sin(\frac{x+y}{2})$

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

Divide (1) and (2),

$\frac{2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})}{-2\sin(\frac{x-y}{2})\sin(\frac{x+y}{2})}=\frac{\frac{4}{5}}{\frac{2}{7}}$

$-\frac{\cot\frac{x+y}{2}}{\tan\frac{x-y}{2}}=\frac{14}{5}$

$-5\cot\frac{x+y}{2}=14\tan\frac{x-y}{2}$

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

Therefore, The value of the expression is $14\tan\frac{x-y}{2}+5\cot\frac{x+y}{2}=0$