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

Options

A. `-(1)/(sqrt(2))`
B. 0
C. `(1)/(sqrt(2))`
D. `(1)/(2 sqrt(2))`

Answer

C

Explanation

The given trigonometric expression is :
`cos[tan^(-1){tan((15pi)/(4))}]`
`=cos[tan^(-1){tan(4pi-(pi)/(4))}]`
`=cos[tan^(-1){-"tan"(pi)/(4)}]=cos[tan^(-1)tan((-pi)/(4))]" "("since" tan^(-1)theta "is defined for"(-pi)/(2)lt theta lt (-pi)/(2))`
`=cos((-pi)/(4))`
`"cos"(pi)/(4)=(1)/(sqrt(2))" "["since cos"(-theta)=cos theta]`