MCQ
Given that $\sin \theta=\frac{a}{b}$, then $\tan \theta$ is equal to:
- A$\frac{b}{\sqrt{a^2+b^2}}$
- B$\frac{b}{\sqrt{b^2-a^2}}$
- C$\frac{a}{\sqrt{a^2-b^2}}$
- ✓$\frac{a}{\sqrt{b^2-a^2}}$
✓
Answer
Correct option: D.
$\frac{a}{\sqrt{b^2-a^2}}$
We have,
$\sin \theta=\frac{a}{b}$
$\sin \theta=\frac{A B}{B C}=\frac{a}{b}($ Given $)$
By Pythagoras theorem, we know,
$B C^2=A B^2+A C^2$
$\Rightarrow b^2=a^2+A C^2$
$\Rightarrow A C^2=b^2-a^2$
$\Rightarrow A C=\sqrt{b^2-a^2}$
Using trigonometric ratios,
$\tan \theta=\frac{A B}{A C}$
$\tan \theta=\frac{a}{\sqrt{b^2-a^2}}$
Hence, the correct option is $(d).$
$\sin \theta=\frac{a}{b}$
$\sin \theta=\frac{A B}{B C}=\frac{a}{b}($ Given $)$
By Pythagoras theorem, we know,
$B C^2=A B^2+A C^2$
$\Rightarrow b^2=a^2+A C^2$
$\Rightarrow A C^2=b^2-a^2$
$\Rightarrow A C=\sqrt{b^2-a^2}$
Using trigonometric ratios,
$\tan \theta=\frac{A B}{A C}$
$\tan \theta=\frac{a}{\sqrt{b^2-a^2}}$
Hence, the correct option is $(d).$
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