यदि `cos A + cos B = m, sin A + sin B = n, ` , जहाँ ` m , n != 0` तो ` sin( A +B) ` कितने के बराबर है ?

Options

A. ` (mn)/(m^(2)+n^(2))`
B. ` (2mn)/(m^(2)+n^(2))`
C. ` (m^(2)+n^(2))/(2mn)`
D. `(mn)/(m+n)`

Answer

B

Explanation

` :' cos A + cos B = m " " `..(i)
तथा ` sin A + sin B = n" "`…(ii)
समी ( i) तथा (ii ) से ,
`(2mn)/(m^(2) +n^(2))`
` = ( 2 ( cos A + cos B) ( sin A + sin B))/((cos A + cos B)^(2) + ( sin A + sin B )^(2))`
`= ( sin 2A + sin 2B + 2 sin ( A + B))/(1+ 1 + 2cos (A - B))`
` = ( sin ( A + B) [ 2 + 2 cos ( A - B)))/( 2 + 2 cos ( A - B))`
` = sin ( A + B)`
` :. sin ( A + B)`
` :. sin ( A + B ) = ( 2 mn)/(m^(2) + n^(2)) `