`tan{(pi)/(4)+1/2cos^(-1)alpha}+tan{(pi)/(4)-1/2cos^(-1)alpha}`
`alpha ne 0` is equal to

Options

A. `alpha`
B. `2alpha`
C. `(2)/(alpha)`
D. none of these

Answer

C

Explanation

We have
`tan{(pi)/(4)+1/2cos^(-1)alpha}+tan{(pi)/(4)-1/2cos^(-1)alpha}`
`tan(pi)/(4)+(theta)/(2)+tan((pi)/(4)-(theta)/(2))` where `theta = cos^(-1)alpha`
`=(1+tan(theta)/(2))/(1-tan(theta)/(2))+(1-tan(theta)/(2))/(1+tan(theta)(theta)/(2))=2(1+tan^(2)(theta)/(2))/(1-tan^(2)(theta)/(2))=(2)/(cos theta)=(2)/(alpha)`