Vectors `vecA` and `vecB` include an angle `theta` between them. If `(vecA + vecB)` and `(vecA- vecB)` respectively subtend angles `alpha and beta` with `vecA`, then `(tan alpha + tan beta)` is

Options

A. `((A sin theta))/((A^(2) +B^(2)cos^(2)theta)) `
B. `((2AB sin theta))/((A^(2) -B^(2)cos^(2)theta)) `
C. `((A^(2) sin^(2) theta))/((A^(2) +B^(2)cos^(2)theta)) `
D. `((B^(2) sin^(2) theta))/((A^(2) -B^(2)cos^(2)theta)) `

Answer

B

Explanation

`tan alpha = (B sin theta)/(A + B cos theta)" "…(i)`
where `alpha ` is the angle made by the vector `(vecA + vecB)` with `vecA`.
Similarly, `tan beta = (B sin theta)/(A - B cos theta) " "…(ii)`
where `beta` is the angle made by the vector `(vecA - vecB)` with `vecA`.
Note that the angle between `vecA and (-vecB)` is `(180^(@) - theta)`.
Adding (i) and (ii), we get
`tan alpha + tan beta = (B sin theta)/(A + b cos theta) + (B sin theta)/(A -Bcostheta)`
` = (AB sin theta - B^(2) sin thetacos theta + AB sin theta + B^(2) sintheta cos theta)/((A + B cos theta)(A-Bcos theta))`
` = (2AB sin theta)/((A^(2) - B^(2) cos ^(2) theta))`