Given that $sin\ \theta = \frac{a}{b}$, then $cos\ \theta$ is equal to
(A) $ \frac{b}{\sqrt{b^{2}-a^{2}}} $
(B) $ \frac{b}{a} $
(C) $ \frac{\sqrt{b^{2}-a^{2}}}{b} $
(D) $ \frac{a}{\sqrt{b^{2}-a^{2}}} $

Given:

$sin\ \theta = \frac{a}{b}$

To do:

We have to find the value of $cos\ \theta$.

Solution:  

We know that,

$sin^2 \theta+cos^2 \theta=1$

$cos \theta=\sqrt{1-sin^2 \theta}$

Therefore,

$cos \theta= \sqrt{1-(\frac{a}{b})^2}$

$=\sqrt{1-\frac{a^2}{b^2}}$

$=\sqrt{\frac{b^2-a^2}{b^2}}$

$=\frac{\sqrt{b^2-a^2}}{b}$