If y = sin ( sin x ) and y '' + f ( x ) · y ' + g ( x ) · y = 0 , then f ( x ) · g ( x ) =

Since $y=\sin \left(\sin x\right) ...\left(i\right)$ 

Differentiating w.r.t. $x$ by chain rule

$y^{\text{'}}=\cos \left(\sin x\right).\cos x ...\left(ii\right)$

Again differentiate w.r.t. $x$

$\Rightarrow y^{\text{'}\text{'}}=-\sin \left(\sin x\right)\times {\cos }^{2}x-\sin x\times \cos \left(\sin x\right)$

$\Rightarrow y^{\text{'}\text{'}}=-\sin \left(\sin x\right)\times {\cos }^{2}x-\sin x\times \cos \left(\sin x\right).\frac{\cos x}{\cos x}$

by equation $\left(i\right)$ and $\left(ii\right)$ 

$\Rightarrow y^{\text{'}\text{'}}=-{\cos }^{2}x.y-\tan x.y^{\text{'}}$

$\Rightarrow y^{\text{'}\text{'}}+\tan x.y^{\text{'}}+{\cos }^{2}x.y=0$

Since we are given that $y^{\text{'}\text{'}}+f\left(x\right)·y^{\text{'}}+g\left(x\right)·y=0$

then by comparison $f\left(x\right)=\tan x, g\left(x\right)={\cos }^{2}x$

Now $f\left(x\right).g\left(x\right)=\tan x.{\cos }^{2}x$

$=\sin x.\cos x.\frac{2}{2}$

Since $2\sin A.\cos A=\sin 2A$

$=\frac{\sin 2x}{2}$.