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Which identity is correct?
A$\cos 2\theta=\dfrac{1+\tan^2\theta}{1-\tan^2\theta}$
B$\sin 2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}$
C$\tan 2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}$
D$\sin 2\theta=2\tan\theta(1+\tan^2\theta)$
Answer & Solution
Correct answer: B. $\sin 2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}$
Using double-angle identities in terms of $\tan\theta$, $\sin 2\theta=\dfrac{2\tan\theta}{1+\tan^2\theta}$. Also, $\cos 2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta}$ and $\tan 2\theta=\dfrac{2\tan\theta}{1-\tan^2\theta}$, so the other options are incorrect.
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