Question:
If $\tan x=-\frac{1}{\sqrt{5}}$ and $\theta$ lies in the IV quadrant, then the value of $\cos x$ is
(a) $\frac{\sqrt{5}}{\sqrt{6}}$
(b) $\frac{2}{\sqrt{6}}$
(c) $\frac{1}{2}$
(d) $\frac{1}{\sqrt{6}}$
Solution:
(a) $\frac{\sqrt{5}}{\sqrt{6}}$
In the fourth quadrant, $\cos x$ and $\sec x$ are positive.
$\cos x=\frac{1}{\sec x}$
$=\frac{1}{\sqrt{\sec ^{2} x}}$
$=\frac{1}{\sqrt{1+\tan ^{2} x}}$
$=\frac{1}{\sqrt{1+\left(-\frac{1}{\sqrt{5}}\right)^{2}}}$
$=\frac{1}{\sqrt{\frac{6}{5}}}$
$=\frac{\sqrt{5}}{\sqrt{6}}$