The value of `c` for which the identity `sin(alpha+beta)sin(alpha-beta)=c-cos^(2)alpha-sin^(2)beta` holds true is

If cos(alpha+beta)=0 then sin^(2)alpha+sin^(2)beta

What is sin (alpha+beta) - 2 sin alpha cos beta + sin(alpha- beta) =

Prove by vector methods that sin(alpha+beta)=sin alpha cos beta+cos alpha sin beta

2sin^(2)beta+4cos(alpha+beta)sin alpha sin beta+cos2(alpha+beta)=

Simplify 2sin^(2)beta+4cos(alpha+beta)sin alpha sin beta+cos2(alpha+beta)

Prove that: cos2 alpha cos2 beta+sin^(2)(alpha-beta)-sin^(2)(alpha+beta)=cos2(alpha+beta)

The value of ((cos alpha+cos beta)/(sin alpha-sin beta))^(n)+((sin alpha+sin beta)/(cos alpha-cos beta))^(n) (where n is a whole number) is equal to

The value of the determinant Delta = |(cos (alpha + beta),- sin (alpha + beta),cos 2 beta),(sin alpha,cos alpha,sin beta),(- cos alpha,sin alpha,- cos beta)| , is

tan (alpha + beta) tan (alpha-beta) = (sin ^ (2) alpha-sin ^ (2) beta) / (cos ^ (2) alpha-sin ^ (2) beta)

If cos alpha+cos beta=0=sin alpha+sin beta then cos2 alpha+cos2 beta=