Question

If $\sin \theta =\frac{a}{b}$, then $\cos \theta =$

• A. $\frac{b}{\sqrt{b^2-a^2}}$
• B. $\frac{\sqrt{b^2-a^2}}{b}$
• C. $\frac{a}{\sqrt{b^2-a^2}}$
• D. $\frac{b}{a}$

Answer 1

The correct option is B $\frac{\sqrt{b^2-a^2}}{b}$

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

$\therefore {\cos }^2\theta =1-{\sin }^2\theta =1-\frac{a^2}{b^2}=\frac{b^2-a^2}{b^2}$

$\Rightarrow \cot \theta =\frac{\sqrt{b^2-a^2}}{b}$.