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tan[pi/4+1/2"Cos"^(-1)x/y]+tan[pi/4-1/2"...

`tan[pi/4+1/2"Cos"^(-1)x/y]+tan[pi/4-1/2"Cos"^(-1)x/y]=`

A

x/y

B

y/x

C

2x/y

D

2y/x

Text Solution

Verified by Experts

The correct Answer is:
D
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