यदि `a = "cos"2alpha + "i sin"2alpha` तथा `b = "cos"2beta + "i sin"2beta` है, तो `sqrt(ab) + (1)/(sqrt(ab))` का मान है

Options

A. `"cos"(alpha + beta)`
B. `2 "cos" (alpha + beta)`
C. `2 "sin" (alpha + beta)`
D. `"sin"(alpha +beta)`

Answer

B

Explanation

`because ab = ("cos" 2alpha + "i sin" 2alpha)("cos" 2beta + "i sin" 2 beta)`
` = "cos"2 (alpha + beta) + "i sin" 2 (alpha + beta)`
`therefore sqrt(ab) = ["cos" 2 (alpha + beta) + "i sin" 2 (alpha + beta)]^(1//2)`
` = "cos"2 * (1)/(2) (alpha + beta) + "i sin" 2 * (1)/(2)(alpha + beta)` [डिमायवर प्रमेय से]
` = "cos" (alpha +beta) + "i sin" (alpha + beta)`
इसी प्रकार, `(1)/(sqrt(ab)) = "cos" (alpha + beta) - "i sin" (alpha + beta)`
`therefore sqrt(ab) + (1)/(sqrt(ab)) = "cos" (alpha + beta) + "i sin" (alpha +beta) + "cos" (alpha + beta) - "i sin" (alpha + beta)`
` = 2 "cos" (alpha +beta)`