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8. Introduction to Trigonometry
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यदि $\cos (\alpha+\beta)=0$ हो, तो $\sin (\alpha-\beta)$ को निम्नलिखित के रूप में बदला जा सकता है:
A
$\cos 2 \beta$
B
$\cos \beta$
C
$\sin \alpha$
D
$\sin 2 \alpha$
Solution
Given, $\cos (\alpha+\beta)=0=\cos 90^{\circ}$ $\left[\because \cos 90^{\circ}=0\right]$
$\Rightarrow \quad \alpha+\beta=90^{\circ}$
$\Rightarrow \quad \alpha=90^{\circ}-\beta$ ……$(i)$
Now, $\sin (\alpha-\beta)=\sin \left(90^{\circ}-\beta-\beta\right)$ [put the value from Eq. $(i)$]
$=\sin \left(90^{\circ}-2 \beta\right)$
$=\cos 2 \beta$ $\left[\because \sin \left(90^{\circ}-\theta\right)=\cos \theta\right]$
Hence, $\sin (\alpha-\beta)$ can be reduced to $\cos 2 \beta$.
Standard 10
Mathematics
