1. ## trig solvein problem

$\sqrt\frac{1+sinA}{1-sinA}=tan[\frac{\pi}{4}+\frac{A}{2}]$ when i simplify finally i get $\frac{1+sinA}{cosa}$then how can i prove it with square root.

2. ## Re: trig solvein problem

$\tan\left(\frac{\pi}{4} + \frac{A}{2}\right)$

$\frac{\tan\left(\frac{\pi}{4}\right) + \tan\left(\frac{A}{2}\right)}{1 - \tan\left(\frac{\pi}{4}\right) \cdot \tan\left(\frac{A}{2}\right)}$

$\frac{1+\tan\left(\frac{A}{2}\right)}{1-\tan\left(\frac{A}{2}\right)}$

$\frac{\cos\left(\frac{A}{2}\right) + \sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2 }\right) - \sin\left(\frac{A}{2}\right)} \cdot \frac{\cos\left(\frac{A}{2}\right) + \sin\left(\frac{A}{2}\right)}{\cos\left(\frac{A}{2 }\right) + \sin\left(\frac{A}{2}\right)}$

$\frac{1 + 2\cos\left(\frac{A}{2}\right)\sin\left(\frac{A}{2} \right)}{\cos^2\left(\frac{A}{2}\right) - \sin^2\left(\frac{A}{2}\right)}$

$\frac{1 + \sin{A}}{\cos{A}}$

3. ## Re: trig solvein problem

i simplify this by the r.h.s and i finally get (1+sina)/cosa then how can i show this as in the l.h.s (in square root)

4. ## Re: trig solvein problem

work backwards ... you figure it out for once.