Question

Prove that
1-Cos A+SinA/1+cosA+sinA=tan(A/2)​

Answer 1

Answer:

hope it helps you.....❤️❤️

Answer 2

Hey there!

Your answer is as follows

Answer with explanation:

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

Previous Knowledge:

1.    $sinA= 2sin(\frac{A}{2} )$

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

3.   $cosA =1- 2sin^{2}(\frac{A}{2})$

4.  $\frac{sin\alpha }{cos \alpha } = tan \alpha$

Proof:

LHS   =>   $\frac{1+sin A-cosA}{1+sinA+cosA}$

Applying (1),(2),(3) we get

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

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

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

Canceling the like terms we get,

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

From (4),

LHS  =  $tan(\frac{A}{2} )$   = RHS

Hence proved!

Thanks!