Question

$\frac{1+sin\theta }{1-sin\theta }={\tan }^2(\frac{\pi }{4}+\frac{\theta }{2})$

• A. True
• B. False

Answer 1

The correct option is A True

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

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

$=\frac{(\cos \frac{\theta }{2}+\sin \frac{\theta }{2}{)}^2}{(\cos \frac{\theta }{2}-\sin \frac{\theta }{2}{)}^2}$

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

$={\tan }^2(\frac{\pi }{4}+\frac{\theta }{2})$

So the relation is $\text{True}$