Question

11) In AABC prove the following: (i) a sin A -b sin B = c sin (A-B)​

Answer 1

Step-by-step explanation:

We know,

$\tt\large\purple{From\:\:laws\:\:of\:\:sine:}$

$\tt \blue{ \frac{sin( A )}{a} =\frac{sin( B )}{b} = \frac{sin( C )}{c} = k }$

$\tt \red{ \implies sin(A) = ka , \: \: sin(B ) = bk \: \:, sin( C) = kc}$

Now,

$\sf \: \color{orange}a \: sin(A) - b \: sin(B)$

$\sf \: = \color{orange} \frac{sin(A)}{k} \cdot\: sin(A) - \frac{sin( B )}{k} \cdot \: sin(B)$

$\sf \: = \color{orange} \frac{1}{k} \{ sin^{2} (A) - sin^{2} (B) \}$

$\sf \: = \color{orange} \frac{1}{k} ( sin (A) - sin (B) )( sin (A) + sin (B))$

$\sf \: = \color{orange} \frac{c}{sin( C) } \cdot2 cos \bigg( \frac{A + B}{2} \bigg) sin \bigg( \frac{A - B}{2} \bigg) \cdot \: 2sin \bigg( \frac{A + B}{2} \bigg) cos \bigg( \frac{A - B}{2} \bigg)$

$\sf \: = \color{orange} \frac{c}{sin( C) } \cdot2 sin \bigg( \frac{A + B}{2} \bigg) cos \bigg( \frac{A + B}{2} \bigg) \cdot \: 2 sin \bigg( \frac{A - B}{2} \bigg) cos \bigg( \frac{A - B}{2} \bigg)$

$\sf \: = \color{orange} \frac{c}{sin( C) } \cdot sin ( A + B ) \cdot \: sin ( A - B)$

$\sf \: = \color{orange} \frac{c}{sin( C) } \cdot sin ( \pi - C ) \cdot \: sin ( A - B)$

$\sf \: = \color{orange} \frac{c}{sin( C) } \cdot sin (C ) \cdot \: sin ( A - B)$

$\sf \: = \color{orange} c \cdot \: sin ( A - B)$

$\sf \: = \color{orange} c \: sin ( A - B)$