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cos[tan^(-1){tan((15pi)/4)}]...

`cos[tan^(-1){tan((15pi)/4)}]`

A

`(1)/sqrt(2)`

B

`-(1)/sqrt(2)`

C

1

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
We have
`cos{tan^(-1)tan(15n)}`
`=cos {tan^(-1)tan(4pi-(pi)/(4))}`
`=cos {tan(tan^(-1))(-pi)/(4)}=cos(-pi)/(4)=cos(pi)/(4)=(1)/sqrt(2)`
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