If `alpha` and `beta` are the two different roots of equations `alpha cos theta+b sin theta=c`, prove that
(a) `tan (alpha-beta)=(2ab)/(a^(2)-b^(2))` (b) `cos(alpha+beta)=(a^(2)-b^(2))/(a^(2)+b^(2))`

Explanation

Since `alpha` and `beta` are the roots of the equation
`alpha cos theta+b sin theta=c,` we have
and `alpha cos beta+b sin beta=c`
Subtracting Eq. ii from Eq. i, we get
`a(cos alpha-cos beta)+b(sin alpha-sinbeta)=0`
or `(b(sin alpha-sin beta)-a(cos beta-cos alpha)=0`
or `2bcos ""(alpha+beta)/(2)sin""(alpha-beta)/(2)=2a sin""(alpha+beta)/(2)sin""(alpha-beta)/(2)`
or `tan""(alpha+beta)/(2)=(b)/(a)` [as `alpha,beta` are different, `sin""(alpha-beta)/(2)ne0`]
Now, `sin (alpha+beta)=(2tan""(alpha+beta)/(2))/(1+tan^(2)""(alpha+beta)/(2))=(2(b)/(a))/(1+(b^(2))/(a^(@)))=(2ab)/(a^(2)+b^(2))`
(a) `tan ""(alpha+beta)=(2tan((alpha+beta)/(2)))/(1-tan^(2)((alpha+beta)/(2))=(2ab)/(a^(2)-b^(2))`
(b) `cos"" (alpha+beta)=(1-tan^(2)((alpha+beta)/(2)))/(1+tan^(2)((alpha+beta)/(2)))=(1-(b^(2))/(a^(2)))/(1+(b^(2))/(a^(2)))=(a^(2)-b^(2))/(a^(2)+b^(2))`