MCQ
If $\tan A = \frac{1}{2},\tan B = \frac{1}{3},$ then $\cos 2A = $
- A$\sin B$
- ✓$\sin 2B$
- C$\sin 3B$
- DNone of these
therefore $2A = 90^\circ - 2B$
$\therefore \cos 2A = \sin 2B$.
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