Prove that: `tan^(-1){pi/4+1/2\ cos^(-1)a/b}+tan{pi/4-1/2\ cos^(-1)a/b}=(2b)/a`

Options

A. `(2a)/(b)`
B. `(2b)/(a)`
C. `(a)/(b)`
D. `(b)/(a)`

Answer

B

Explanation

`tan[(pi)/(4)+(1)/(2)"cos"^(-1)((a)/(b))]+tan[(pi)/(4)-(1)/(2)"cos"^(-1)((a)/(b))]`
`=tan[(pi)/(4)+phi]+tan[(pi)/(4)-phi]`
`["put"(1)/(2)"cos"^(-1)((a)/(b))=phirArrcos2phi=(a)/(b)]`
`=(1+tanphi)/(1-tanphi)+(1-tanphi)/(1+tanphi)`
`=(2(1+tan^(2)phi))/(1-tan^(2)phi)`
`=(2sec^(2)phi)/((cos^(2)phi-sin^(2)phi)/(cos^(2)phi))`
`=(2)/(cos2phi)=(2b)/(a)`