If `1/2sin^(-1)[(3sin2theta)/(5+4cos2theta)]=tan^(-1)x ,t h e nx=` (a)`tan3theta` (b) `3tantheta` (c) `(1/3)tantheta` (d) `3cottheta`

Explanation

`1/2sin^-1[(3sin2theta)/(5+4cos2theta)] = tan^-1x`

`=>sin^-1[(3sin2theta)/(5+4cos2theta)] = 2tan^-1 x`

`=>sin^-1[(3((2tantheta)/(1+tan^2theta)))/(5+4((1-tan^2theta)/(1+tan^2theta)))] = 2tan^-1 x`

`=>sin^-1[(6tantheta)/(5+5tan^2theta +4 - 4tan^2theta)] = sin^1((2x)/(1+x^2))`

`=>sin^-1[(6tantheta)/(9+tan^2theta)] = sin^1((2x)/(1+x^2))`

`=>sin^-1[(2(tantheta/3))/(1+(tantheta/3)^2)] = sin^1((2x)/(1+x^2))`

`=> x = 1/3(tantheta)`

So, option `c` is the correct option.