Question

If $\mathrm{sin\theta }=\frac{\mathrm{a}}{\mathrm{b}}$then $\mathrm{cos\theta }$ is equal to.

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

Answer 1

The correct option is C

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

Explanation for the correct Option:

Find the required value.

Put the Values in the Identity

$$\begin{gathered} {\cos }^2\theta =1-{\sin }^2\theta [\because {\sin }^2\theta +{\cos }^2\theta =1] \\ \Rightarrow {\cos }^2\theta =1-{\left(\raisebox{1ex}{$\mathrm{a}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{b}$}\right.\right)}^2[\sin \mathrm{\theta }=\frac{\mathrm{a}}{\mathrm{b}}] \\ \Rightarrow \mathrm{cos\theta }=\frac{\sqrt{{\mathrm{b}}^2-{\mathrm{a}}^2}}{\mathrm{b}} \end{gathered}$$

Hence, option $\mathrm{C}$ is the correct answer.