ज्ञात करें : `tan[(1)/(2) cos^(-1)""(sqrt(5))/(3)]`

Options

A. `(3-sqrt5)/2`
B. `(3+sqrt5)/2`
C. `1`
D. `sqrt3`

Answer

A

Explanation

माना कि `cos^(-1)((sqrt(5))/(2))= 2theta` तो `cos 2theta = (sqrt(5))/(3)` तथा `0 le 2theta le pi`
अब, `cos 2theta = (sqrt(5))/(3)`
`therefore (1-tan^(2)theta)/(1+tan^(2)theta) = (sqrt(5))/(3)` या, `(1+tan^(2)theta)/(1-tan^(2)theta) = (3)/(sqrt(5))`
या, `(2tan^(2)theta)/(2) = (3-sqrt(5))/(3+sqrt(5)) " "` [योगान्तर निष्पत्ति से]
या, `tan^(2) = ((3-sqrt(5))/(3+sqrt(5)))((3-sqrt(5))/(3-sqrt(5)))" "...(i)`
लेकिन `0 le 2theta le pi " " therefore le theta le (pi)/(2)`
`rArr theta` पहले पाद में है | अतः `tan theta` धन होगा |
`therefore` (1) से, `tan theta = (3-sqrt(5))/(2)`