Question:
If $\sin \theta=\frac{3}{5}$, 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) $\frac{a}{\sqrt{b^{2}-a^{2}}}$
Solution:
(c) Given, $\sin \theta=\frac{a}{b}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1 \Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta}\right]$
$\therefore \quad \cos \theta=\sqrt{1-\sin ^{2} \theta}$
$=\sqrt{1-\left(\frac{a}{b}\right)^{2}}=\sqrt{1-\frac{a^{2}}{b^{2}}}=\frac{\sqrt{b^{2}-a^{2}}}{b}$
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