If `3(cos 2phi-cos2theta)=1-cos2phi cos2theta`, then `tan theta= k tan phi`, where `theta, phiin (0,(pi)/(2))`, where k =

Options

A. `sqrt(2)`
B. `(1)/(sqrt(2))`
C. `sqrt(3)`
D. `(1)/(sqrt(3))`

Answer

A

Explanation

`3 cos 2phi-3 cos 2theta =1-cos 2phi cos 2 theta`
`rArr (3+cos 2theta) cos 2phi =1+3 cos 2theta`
`rArr cos 2 phi = (1+3 cos 2theta)/(3+cos 2theta)`
`rArr (1-cos 2phi)/(1+cos 2phi)=(1)/(2).(1-cos 2theta)/(1+cos 2theta)` (using componendo and dividendo)
`rArr tan^(2) phi = (1)/(2) tan^(2) theta`
`rArr tan theta = sqrt(2) tan phi`
`rArr k = sqrt(2)`