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If $\sin \alpha = \frac{\sqrt{3}}{2}$, $\cos \beta = \frac{\sqrt{3}}{2}$, then $\tan \alpha \cdot \tan \beta$ is:
- (a)$\sqrt{3}$
- (b)$1$
- (c)$\frac{1}{\sqrt{3}}$
- (d)$0$
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Sol.
(C) $1$
(C) $1$
CBSE Class 10 Maths PYQ · Trigonometry · Find Angle of T-Ratio · 1 Mark · March 2024 · Standard
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