If $f(x) $ and g(x) are functions such that $f(x+y) =f(x)g(y) +g(x) f(y) $ then ... Question : 
If $f(x) $ and g(x) are functions such that $f(x+y) =f(x)g(y) +g(x) f(y) $ then $\begin{vmatrix}
f(\alpha) & g(\alpha) & f(\alpha + \theta) \\ f(\beta) & g(\beta) & f(\beta +\theta) \\ f(\gamma) & g(\gamma) & f(\gamma + \theta) \\
\end{vmatrix}$ is independent of 
(a) $\alpha$ 
(b) $\beta$
(c) $\gamma$ 
(d) $\theta$ 
My approach  : Let x =0, y =0 the given equation becomes : 
$f(0+0) = f(0) g(0) +g(0)f(0) $ 
$\Rightarrow f(0) =2f(0)g(0) $ 
$\Rightarrow f(0) \{ 1-2g(0)\} =0$ 
$\Rightarrow f(0) =0; g(0) =\frac{1}{2}$ 
Now assuming x=1, y=0 
we get $f(1) =\frac{1}{2}$ 
But I dont know how to proceed further to get f(x) and g(x) please guide thanks.
 A: Note that
$f(\alpha + \theta) = g(\theta)f(\alpha) + f(\theta)g(\alpha), \tag{1}$
$f(\beta + \theta) = g(\theta)f(\beta) + f(\theta)g(\beta), \tag{2}$
$f(\gamma + \theta) = g(\theta)f(\gamma) + f(\theta)g(\gamma). \tag{3}$
(1)-(3) may be interpreted as asserting that the third column of 
$\begin{vmatrix}
f(\alpha) & g(\alpha) & f(\alpha + \theta) \\ f(\beta) & g(\beta) & f(\beta +\theta) \\ f(\gamma) & g(\gamma) & f(\gamma + \theta) \\
\end{vmatrix} \tag{4}$
is a linear combination of the first two columns; as such, the determinant (4) vanishes for all values of $\alpha, \beta, \gamma, \theta$.
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
A: HINT:
We have $f(\alpha+\theta)=f(\alpha)g(\theta)+g(\alpha)f(\theta)$
Apply $$C_3'=C_3-[C_1\cdot g(\theta)+C_2\cdot f(\theta)]$$ where $C_i$ is the $i$th column $1\le i\le3$
A: It is (d).
Since $f(\alpha+\theta)=f(\alpha)g(\theta)+f(\theta)g(\alpha)$,$f(\beta+\theta)=f(\beta)g(\theta)+f(\theta)g(\beta)$  and $f(\gamma +\theta)=f(\gamma)g(\theta)+f(\theta)g(\gamma)$,then it becomes 
$\begin{vmatrix}
f(\alpha) & g(\alpha) & 0 \\ f(\beta) & g(\beta) & 0 \\ f(\gamma) & g(\gamma) & 0\\
\end{vmatrix}$
