If cosx + cosy = 4
5
and cosx - cosy = 2
7
, find the value of
x - y x + y 14tan + 5cot
2 2
.
Answers
Answer:
14\tan\frac{x-y}{2}+5\cot\frac{x+y}{2}=014tan
2
x−y
+5cot
2
x+y
=0
Step-by-step explanation:
Given : If \cos x+\cos y=\frac{4}{5}cosx+cosy=
5
4
and \cos x-\cos y=\frac{2}{7}cosx−cosy=
7
2
To find : The value of 14\tan(\frac{x-y}{2})+5\cot\frac{x+y}{2}14tan(
2
x−y
)+5cot
2
x+y
?
Solution :
If \cos x+\cos y=\frac{4}{5}cosx+cosy=
5
4
and \cos x-\cos y=\frac{2}{7}cosx−cosy=
7
2
Applying trigonometric identities,
\cos x+\cos y=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})cosx+cosy=2cos(
2
x+y
)cos(
2
x−y
)
\frac{4}{5}=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})
5
4
=2cos(
2
x+y
)cos(
2
x−y
) .....(1)
\cos x-\cos y=-2\sin(\frac{x-y}{2})\sin(\frac{x+y}{2})cosx−cosy=−2sin(
2
x−y
)sin(
2
x+y
)
\frac{2}{7}=-2\sin(\frac{x-y}{2})\sin(\frac{x+y}{2})
7
2
=−2sin(
2
x−y
)sin(
2
x+y
) .....(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}}
−2sin(
2
x−y
)sin(
2
x+y
)
2cos(
2
x+y
)cos(
2
x−y
)
=
7
2
5
4
-\frac{\cot\frac{x+y}{2}}{\tan\frac{x-y}{2}}=\frac{14}{5}−
tan
2
x−y
cot
2
x+y
=
5
14
-5\cot\frac{x+y}{2}=14\tan\frac{x-y}{2}−5cot
2
x+y
=14tan
2
x−y
14\tan\frac{x-y}{2}+5\cot\frac{x+y}{2}=014tan
2
x−y
+5cot
2
x+y
=0
Therefore, The value of the expression is 14\tan\frac{x-y}{2}+5\cot\frac{x+y}{2}=014tan
2
x−y
+5cot
2
x+y
=0