Question

M=COSA/COSB. .N=SINA/SINB THEN (M^2+N^2)SIN^2B=?

Answer 1

)(m2-n2)sin2B=sin2Asin2B-cos2Acos2Bsin2B=sin2Acos2B-cos2Asin2Bsin2Bcos2Bsin2B=sin2Acos2B-cos2Asin2Bcos2B=(1-cos2A)cos2B-cos2A(1-cos2B)cos2B=cos2B-cos2Acos2B-cos2A+cos2Acos2Bcos2B=cos2B-cos2Acos2B=1-cos2Acos2B=1-n2Proved(2) (n2-m2)sin2A=cos2Acos2B-sin2Asin2Bsin2A=cos2Asin2B-sin2Acos2Bcos2Bsin2Bsin2A=sin2Asin2Bcos2Asin2B-sin2Acos2Bcos2B=sin2Asin2Bcos2A(1-cos2B)-(1-cos2A)cos2Bcos2B=sin2Asin2Bcos2A-cos2Acos2B-cos2B+cos2Acos2Bcos2B=sin2Asin2Bcos2A-cos2Bcos2B=sin2Asin2Bcos2Acos2B-cos2Bcos2B=sin2Asin2Bcos2Acos2B-1=m2(n2-1)