Question

prove that sinA+sinB/cosA-cosB=tan(A+B)/2​

Answer 1

Answer:

Step-by-step explanation:

Correct Question :-

Prove that :-

$\dfrac{sinA+sinB}{cosA+cosB}=tan\left(\dfrac{A+B}{2}\right)$

To Prove :-

$\dfrac{sinA+sinB}{cosA+cosB}=tan\left(\dfrac{A+B}{2}\right)$

Formula Required :-

sinx + siny = 2sin(x+y/2)cos(x-y/2)

cosx + cosy = 2cos(x+y/2)cos(x-y/2)

Solution :-

Taking L.H.S :-

$=\dfrac{sinA+sinB}{cosA+cosB}$

Substituting the formula :-

$=\dfrac{2sin\left(\dfrac{A+B}{2}\right)cos\left(\dfrac{A-B}{2}\right)}{2cos\left(\dfrac{A+B}{2}\right)cos\left(\dfrac{A-B}{2}\right)}$

Cancelling the common terms :-

$=\dfrac{sin\left(\dfrac{A+B}{2}\right)}{cos\left(\dfrac{A+B}{2}\right)}$

= tan(A+B/2)

= R.H.S

Hence Proved