Prove ρ(X,Z) = ρ

The covariance of two random variables $$X$$ and $$Y$$ is defined by: $$\mathrm{Cov}(X,Y）= \operatorname{E}(X-\operatorname{E}(X))(Y-\operatorname{E}(Y))=\operatorname{E}(XY)-\operatorname{E}(X)\operatorname{E}(Y)$$

Another related definition is correlation coefficient $$\rho(X,Y) = \frac{\mathrm{Cov}(X,Y)}{\sqrt{\mathrm{Var}(X)\mathrm{Var}(Y)}}$$

It can be proved that the correlation coefficient $$\rho(X,Y)$$ always lies between −1 and +1. $$X$$ and $$Y$$ are two independent standard normal random variables. We now define another random variable $$Z$$ by $$Z=\rho X+\sqrt{1-\rho^2}\cdot Y$$ where $$\rho \in [−1,1]$$.

How can one prove hat $$\rho(X,Z) = \rho$$?

as per definition $$E[Y]=E[X]=0$$, and also $$\mathit{Var}(X)=\mathit{Var}(Y)=1$$.
$$\mathit{Var}(Z)=\mathit{Var}(\rho X+\sqrt{1-\rho^2}Y)=E\big[\big(\rho X+\sqrt{1-\rho^2}Y-\rho \langle X \rangle+\sqrt{1-\rho^2} \langle Y \rangle \big)^2\big]=E\big[\big(\rho X+\sqrt{1-\rho^2}Y \big)^2\big]=E\big[\rho^2 X^2+2\rho \sqrt{1-\rho^2}XY+(1-\rho^2)Y^2\big]=\rho^2E\big[X^2\big]+2\rho\sqrt{1-\rho^2}E\big[XY\big]+(1-\rho^2)E\big[Y^2\big]=1$$
Therefore $$\rho(X,Z)=\mathit{Cov}(X,Z)$$. Further, also as per definition $$E[XY]=0$$. Applying the formula for the covariance:
$$\mathit{Cov}(X,Z)=\mathit{Cov}(X,\rho X+\sqrt{1-\rho^2}Y)=E\big[(X-\langle X \rangle)(\rho X+\sqrt{1-\rho^2}Y-\rho \langle X \rangle-\sqrt{1-\rho^2}\langle Y \rangle)\big]=E\big[X(\rho X+\sqrt{1-\rho^2}Y)\big]=\rho E[X^2]+\sqrt{1-\rho^2}E[XY]=\rho E[X^2]=\rho$$
The covariance operator is linear and since $$X$$ and $$Y$$ are independent any covariance between them is zero. Therefore, the only part that matters is the X-part of the covariance:
$$Cov[X, Z] = \rho Cov[X, X] + Cov[..., ...Y] = \rho Var[X] = \rho$$.