Question

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

Answer 1

Answer:

$\frac{sin(A)-sin(B)}{cos(A)+cos(B)} = tan(\frac{A-B}{2} )$

Step-by-step explanation:

We know that

Then,

dividing above equations

$sin(A) - sin(B) = 2cos(\frac{A+B}{2})sin(\frac{A-B}{2} ) \\ cos(A)+cos(B) = 2cos(\frac{A+B}{2})\frac{sin(A) - sin(B) }{cos(A)+cos(B}=\frac{2cos(\frac{A+B}{2})sin(\frac{A-B}{2} )}{2cos(\frac{A+B}{2})cos(\frac{A-B}{2})} \\ \\ => \frac{sin(\frac{A-B}{2})}{cos(\frac{A-B}{2})} \\ \\=>tan(\frac{A-B}{2})$

Answer 2

SinA+SinB

------------------ =Tan(A+B)/2

CosA+CosB