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8. Introduction to Trigonometry
easy
यदि $\sin \theta=\frac{a}{b}$ दिया है, तो $\cos \theta$ बराबर है:
A
$\frac{b}{\sqrt{b^{2}-a^{2}}}$
B
$\frac{b}{a}$
C
$\frac{a}{\sqrt{b^{2}-a^{2}}}$
D
$\frac{\sqrt{b^{2}-a^{2}}}{b}$
Solution
Given, $\quad \sin \theta=\frac{a}{b}$ $\left[\because \sin ^{2} \theta+\cos ^{2} \theta=1 \Rightarrow \cos \theta=\sqrt{1-\sin ^{2} \theta}\right]$
$\cos \theta=\sqrt{1-\sin ^{2} \theta}$
$=\sqrt{1-\left(\frac{a}{b}\right)^{2}}$
$=\sqrt{1-\frac{a^{2}}{b^{2}}}$
$=\frac{\sqrt{b^{2}-a^{2}}}{b}$
Standard 10
Mathematics
