MCQ
Given that $\sin \theta=\frac{a}{b}$, then $\cos \theta$ is equal to
- A$\frac{b}{\sqrt{b^2-a^2}}$
- B$\frac{b}{a}$
- ✓$\frac{\sqrt{b^2-a^2}}{b}$
- D$\frac{a}{\sqrt{b^2-a^2}}$
✓
Answer
Correct option: C.
$\frac{\sqrt{b^2-a^2}}{b}$
Given $\sin \theta=\frac{a}{b}$
$\because \cos \theta=\sqrt{1-\sin ^2 \theta} \ \left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$
$\Rightarrow \cos \theta=\sqrt{1-\left(\frac{a}{b}\right)^2}$
$=\sqrt{1-\frac{a^2}{b^2}}=\frac{\sqrt{b^2-a^2}}{b}$
$\because \cos \theta=\sqrt{1-\sin ^2 \theta} \ \left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$
$\Rightarrow \cos \theta=\sqrt{1-\left(\frac{a}{b}\right)^2}$
$=\sqrt{1-\frac{a^2}{b^2}}=\frac{\sqrt{b^2-a^2}}{b}$
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