Question

$\sin 4\alpha =4\tan \alpha \frac{1-{\tan }^2\alpha }{{\left(1+{\tan }^2\alpha \right)}^2}$

• A. True
• B. False

Answer 1

The correct option is A True

As we know that

$\sin 2\theta =\frac{2\tan \theta }{1+{\tan }^2\theta }$ and $\cos 2\theta =\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }$

$\sin 4\alpha =2\sin 2\alpha \cos 2\alpha$ $(\because \sin 2\theta =2\sin \theta \cos \theta )$

$=2\times \frac{2\tan \alpha }{1+{\tan }^2\alpha }\times \frac{1-{\tan }^2\alpha }{1+{\tan }^2\alpha }$

$=4\tan \alpha \frac{1-{\tan }^2\alpha }{(1+{\tan }^2\alpha {)}^2}$

So the given relation is $\text{True}$