यदि `x = cos alpha + i sin alpha, y = cos beta + i sin beta`, तो सिद्ध कीजिए
`(x -y)/(x +y) =i "tan"(alpha - beta)/(2)`

Explanation

यदि `x = cos alpha + i sin alpha, y = cos beta + i sin beta`, तो
बायां पक्ष `= (x -y)/(x +y)`
`= (cos alpha + i sin alpha - cos beta - i sin beta)/(cos alpha + i sin alpha + cos beta + i sin beta)`
`= ([cos alpha - cos beta + i sin alpha - sin beta])/([cos alpha + cos beta + i sin alpha + sin beta])`
`= (-2 "sin"(alpha + beta)/(2) "sin"(alpha - beta)/(2) + i.2 "cos"(alpha + beta)/(2) "sin"(alpha - beta)/(2))/(2"cos"(alpha + beta)/(2) "cos"(alpha - beta)/(2) + i.2 "sin"(alpha+ beta)/(2) ."cos"(alpha - beta)/(2))`
`= (2"sin"(alpha - beta)/(2)[-"sin"(alpha + beta)/(2) + i "cos"(alpha + beta)/(2)])/(2"cos"(alpha - beta)/(2) ["cos"(alpha + beta)/(2) + i "sin"(alpha + beta)/(2)])`
`= (i "sin"(alpha - beta)/(2) ["cos"(alpha + beta)/(2) + i "sin"(alpha + beta)/(2)])/("cos"(alpha - beta)/(2)["cos"(alpha + beta)/(2) + i "sin"(alpha + beta)/(2)])`
`=i "tan"(alpha - beta)/(2)` =बायां पक्ष