Question

x/y=sinA/sinB then x. cotA+y.cot B /x+y​

Answer 1

Answer:

$\frac{x.cotA+y.cotB}{x+y}=cot(\frac{A+B}{2})$

Step-by-step explanation:

Given:

$\frac{x}{y}=\frac{sinA}{sinB}$

$\implies\:\frac{x}{sinA}=\frac{y}{sinB}=k(say)$

$\implies\:x=k\:sinA$

$y=k\:sinB$

Now,

$\frac{x.cotA+y.cotB}{x+y}$

$=\frac{x.\frac{cosA}{sinA}+y.\frac{cosB}{sinB}}{k\:sinA+k\:sinB}$

$=\frac{k.cosA+k.cosB}{k(sinA+sinB)}$

$=\frac{k(cosA+cosB)}{k(sinA+sinB)}$

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

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

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

$\implies\frac{x.cotA+y.cotB}{x+y}=cot(\frac{A+B}{2})$