Question

sin (B-C) / sin B. sin C + sin (C-A) / sin C. sin A + sin (A-B) / sin A. sin B = 0​

Answer 1

Answer:

proved

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\frac{sin(B-C)}{sinB.sinC}$  +  $\frac{sin(C-A)}{sinC.sinA}$  +  $\frac{sin(A-B)}{sinA.sinB}$

= $\frac{sinB.cosC - cosB.sinC}{sinB.sinC}$ + $\frac{sinC.cosA - cosC.sinA}{sinC. sinA}$ + $\frac{sinA.cosB - cosA.sinB}{sinA. sinB}$

= $\frac{sinB.cosC}{sinB.sinC} - \frac{cosB.sinC}{sinB.sinC}$ + $\frac{sinC.cosA}{sinC. sinA} - \frac{cosC.sinA}{sinC. sinA}$ + $\frac{sinA.cosB}{sinA. sinB} - \frac{cosA.sinB}{sinA. sinB}$

= $\frac{cosC}{sinC} - \frac{cosB}{sinB}$ + $\frac{cosA}{sinA} - \frac{cosC}{sinC}$ + $\frac{cosB}{sinB} - \frac{cosA}{sinA}$

= $cotC - cotB$ + $cotA - cotC$ + $cotB - cotA$

= 0

=> $\frac{sin(B-C)}{sinB.sinC}$  +  $\frac{sin(C-A)}{sinC.sinA}$  +  $\frac{sin(A-B)}{sinA.sinB}$$\frac{sin(B-C)}{sinB.sinC}$  = 0

Hence proved