Question

Question 4
If $sin\theta =\frac{a}{b}$, 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) $\frac{a}{\sqrt{b^2-a^2}}$

Answer 1

The answer is (C).
Given, $sin\theta =\frac{a}{b}\lfloor \because si{n}^2\theta +co{s}^2\theta =1\Rightarrow cos\theta =\sqrt{1-si{n}^2\theta }\rfloor$

$\therefore cos\theta =\sqrt{1-si{n}^2\theta }$

$=\sqrt{1-{\left(\frac{a}{b}\right)}^2}=\sqrt{1-\frac{a^2}{b^2}}=\frac{\sqrt{b^2-a^2}}{b}$