Proving that the $\frac {\xi +\zeta\eta}{\sqrt {1+\zeta^2}}$ has normal distribution (0,1) The task: $\xi, \eta, \zeta \sim N(0,1)$ and independent. Prove, that $\frac {\xi +\zeta\eta}{\sqrt {1+\zeta^2}} \sim N(0,1).$ (1)
It is clear, that with fixed $\zeta$ we get, that (1) has expected value = 0 (as the sum of normal distributed values) and variance = 1 (as the sum of $(\frac {1}{\sqrt {1+\zeta^2}})^2$ and $(\frac {\zeta}{\sqrt {1+\zeta^2}})^2$). And what to do with un-fixed value I don't know. There was a small tip -imagine, that $\zeta$ is discrete value (for example getting 3 different values) and use the full probability formula $(P(B)=\sum P(B|A_{j})P(A_{j}))$. 

Let $R=\frac{\xi+\zeta\eta}{\sqrt{1+\zeta^2}}$. You want to show $E\exp(itR)=\exp(-t^2/2)$.  Write $E\exp(itR)=E(E[\exp(itR)|\zeta])$.  The inner, or conditional expectation is $$\begin{align*}E[\exp(itR)|\zeta]&=\tag{*}
E\exp(it\xi/\sqrt{1+\zeta})|\zeta) \times E\exp(it\eta/\sqrt{1+\zeta})|\zeta)\\
&= \exp(-\frac{t^2}{2(1+\zeta^2)}) \exp (-\frac{t^2\zeta^2}{2(1+\zeta^2)})\\
&= \exp(-\frac{t^2}2).
\end{align*}$$
The first step, at (*), is because $\xi$ and $\eta$ are conditionally independent given $\zeta$.
So the outer expectation is also $\exp(-t^2/2)$.
