Correct Answer - 2
We have
`" " ( sin ^(3) theta )/( sin ( 2theta + alpha )) = ( cos ^(3) theta )/( cos( 2theta + alpha )) = k` ( let )
`rArr ( sin ^(4)theta )/( sin theta sin (2 theta + alpha ))= ( cos ^(4)theta )/( cos theta cos ( 2theta + alpha )) = k`
`rArr cos ^(4) theta - sin ^(4) theta = k[cos theta cos ( 2theta + alpha) - sin theta sin ( 2 theta + alpha)]`
`rArr cos 2 theta = k cos ( 3 theta + alpha )" "` ... (1)
Also,
`" " ( sin ^(3)theta cos theta ) /( sin ( 2theta + alpha ) cos theta) = ( sin theta cos ^(3) theta)/( sin theta cos ( 2theta + alpha )) =k`
`rArr sin ^(3) theta cos theta + sin theta cos^(3) theta `
` " " k ( sin ( 2theta + alpha ) cos theta + sin theta cos ( 2 theta + alpha ))`
`rArr sin theta cos theta ( sin ^(2) theta + cos ^(2) theta ) =k sin ( 3 theta + alpha)`
`rArr sin 2 theta = 2k sin ( 3 theta + alpha )" "` ... (2)
From (1) and (2), we get
`" " tan 2 theta = 2 tan ( 3theta + alpha ) `
