Question

proof that 1+sinA -cosA /1+sinA+cosA = tanA/2

Answer 1

Sina=2sin(a/2) cos(a/2)

Cosa=1-sin^2(a/2)

Cosa=2cos(a/2) -1

Answer 2

Answer and Explanation:

To prove : $\frac{1+\sin A-\cos A}{1+\sin A+\cos A}=\tan (\frac{A}{2})$

Proof :

Taking LHS,      

$LHS=\frac{1+\sin A-\cos A}{1+\sin A+\cos A}$

Using the trigonometry identities we can write,

1) $\sin A=2\sin (\frac{A}{2})\cos (\frac{A}{2})$

2) $\cos A=2\cos^2(\frac{A}{2})-1$

3) $\cos A=1-2\sin^2(\frac{A}{2})$

Applying these identities,

$LHS=\frac{1+2\sin (\frac{A}{2})\cos (\frac{A}{2})-1+2\sin^2(\frac{A}{2})}{1+2\sin (\frac{A}{2})\cos (\frac{A}{2})+2\cos^2(\frac{A}{2})-1}$

$LHS=\frac{2\sin (\frac{A}{2})\cos (\frac{A}{2})+2\sin^2(\frac{A}{2})}{2\sin (\frac{A}{2})\cos (\frac{A}{2})+2\cos^2(\frac{A}{2})}$

$LHS=\frac{2\sin (\frac{A}{2})(\cos (\frac{A}{2})+\sin(\frac{A}{2}))}{2\cos (\frac{A}{2})(\sin (\frac{A}{2})+\cos(\frac{A}{2}))}$

Cancel the like term,

$LHS=\frac{\sin (\frac{A}{2})}{\cos (\frac{A}{2})}$

$LHS=\tan \frac{A}{2}$

$LHS=RHS$

Hence proved.