A line OA of length r starts from its initial position OX and traces an angle AOB = `alpha` in the anitclockwise direction. It then traces back in the clockwise direction an angle BOC = `3theta` ( where `alpha gt 3 theta`). L is the foot of the perpendicular from C on OA. Also, `(sin^(3)theta)/(CL) = (cos^(3)theta)/(OL)=1`
`(1-rcos alpha)/(r sin alpha)` is equal to

Options

A. `tan^(2) theta`
B. `cot^(2) theta`
C. `cot 2theta`
D. `tan 2 theta`

Answer

D

Explanation

`OL = rcos (alpha - 3theta)`
`CL = r sin(alpha - 3theta)`

So, `(sin^(3)theta)/(rsin(alpha - 3theta)) = ( cos^(3) theta)/(r cos (alpha -3theta)) =1 `
`rArr (sin^(3)theta)/( sin (alpha - 3theta)) = (cos^(3) theta )/(cos(alpha - 3theta)) = r`
Now `cos^(4) theta - sin^(4) theta `
`" " = costheta (r cos( alpha - 3theta))- sin theta ( rsin ( alpha - 3theta))`
`rArr cos 2theta = r cos (alpha - 2theta)`
`rArr cos 2theta = r (cos alpha cos 2theta + sin alpha sin 2theta)`
`rArr (1-r cos alpha) cos 2 theta = r sin alpha sin 2theta`
`rArr (1-rcos alpha) cos 2theta = rsin alpha sin 2theta" " ` ...(1)
`cos^(3)theta sin theta + sin^(3)theta cos theta `
`= r [cos( alpha - 3theta ) sin theta + sin (alpha - 3theta) cos theta ]`
`rArr sin theta cos theta = r sin (alpha - 2theta)`
`rArr sin 2theta = 2r [ sin alpha cos 2theta - cos alpha sin2 theta]`
`rArr ( 1+ 2r cos alpha) sin 2theta = 2r sin alpha cos 2 theta`
`rArr (1+ 2r cos alpha )/( 2r sin alpha) = cot 2 theta " "`...(2)
From (1) and (2), we get
`" " (1-r cos alpha )/(rsin alpha) = (2r sin alpha)/(1+2rcos alpha)`
`rArr 1-r cos alpha + 2r cos alpha - 2r^(2) cos^(2) alpha = 2r^(2) sin^(2) alpha`
`rArr (2r^(2) -1)/(r) = cos alpha`