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sin alpha + sin beta = a "and" cos alpha...

`sin alpha + sin beta = a "and" cos alpha + cos beta = b`
`sin (alpha + beta) =`

A

`(2ab)/(a^(2) + b^(2))`

B

`(2ab)/(a^(2) - b^(2))`

C

`(a^(2) + b^(2))/(2ab)`

D

`(a^(2) - b^(2))/(2ab)`

Text Solution

Verified by Experts

The correct Answer is:
A
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Knowledge Check

  • sin alpha + sin beta = a "and" cos alpha + cos beta = b cos (alpha + beta) =

    A
    `(a^(2)-b^(2))/(a^(2)+b^(2))`
    B
    `-(a^(2) - b^(2))/(a^(2) + b^(2))`
    C
    `(a^(2) + b^(2))/(a^(2) - b^(2))`
    D
    `-(a^(2) + b^(2))/(a^(2) - b^(2))`

  • sin alpha + sin beta = a "and" cos alpha + cos beta = b (tan alpha + tan beta)/(1-tan alpha tan beta)=

    A
    `(a^(2) + b^(2))/(2ab)`
    B
    `(b^(2)-a^(2))/(2ab)`
    C
    `(2ab)/(b^(2) + a^(2))`
    D
    `(2ab)/(b^(2) - a^(2))`

  • If sin alpha = sin beta and cos alpha = cos beta then-

    A
    `sin ""(alpha + beta)/(2)=0`
    B
    `cos"" (alpha -beta)/(2)=0`
    C
    `cos ""(alpha+beta)/(2) =0`
    D
    `sin ""(alpha -beta)/(2)=0`

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