If sin x

We have :

$\sin x+\cos x=0$

$\Rightarrow \sin x=-\cos x$

$\Rightarrow \frac{\sin x}{\cos x}=-1$

$\Rightarrow \tan x=-1$

Now, $x$ is in the fourth quadrant.

In the fourth quadrant, $\cos x$ and $\sec x$ are positive and all the other four $\mathrm{T}$-ratios are negative.

$\therefore \sec x=\sqrt{1+\tan ^{2} x}=\sqrt{1+(-1)^{2}}=\sqrt{2}$

$\cos x=\frac{1}{\sec x}=\frac{1}{\sqrt{2}}$

And, $\sin x=-\sqrt{1-\cos ^{2} x}=-\sqrt{1-\left(\frac{1}{\sqrt{2}}\right)^{2}}=\frac{-1}{\sqrt{2}}$

$\therefore \quad \sin x=\frac{-1}{\sqrt{2}}$ and $\cos x=\frac{1}{\sqrt{2}}$