If `cos (A+B) = alpha cos A cos B + beta sin A sin B,` then `(alpha, beta)` is

If x cos alpha + y sin alpha = x cos beta + y sin beta "then" (sin alpha - cos alpha - sin beta + cos beta)/(sin alpha + cos alpha - sin beta - cos beta)=

If cos alpha+cos beta=0=sin alpha+sin beta, then cos2 alpha+cos2 beta is equal to (a)-2sin(alpha+beta)(b)-2cos(alpha+beta)(c)2sin(alpha+beta)(d)2cos(alpha+beta)

If cos alpha+cos beta=0=sin alpha+sin beta, then cos2 alpha+cos2 beta is equal to

(i) sin (alpha + beta) = sin alpha cos beta + cos alpha sin beta) (ii) sin (alpha-beta)= sin alpha cos beta- cos alpha sin beta Proof

Evaluate {:[( cos alpha cos beta , cos alpha sin beta , -sin alpha ),( -sin beta , cos beta, 0),( sin alpha cos beta, sin alpha sin beta, cos alpha ) ]:} =0

Evaluate {:|( cos alpha cos beta , cos alpha sin beta , -sin alpha ),( -sin beta , cos beta, 0),( sin alpha cos beta, sin alpha sin beta, cos alpha ) |:} =0

Evaluate {:|( cos alpha cos beta , cos alpha sin beta , -sin alpha ),( -sin beta , cos beta, 0),( sin alpha cos beta, sin alpha sin beta, cos alpha ) |:} =1

Evaluate {:[( cos alpha cos beta , cos alpha sin beta , -sin alpha ),( -sin beta , cos beta, 0),( sin alpha cos beta, sin alpha sin beta, cos alpha ) ]:} =0