If cos A/cosB= m andCOS/
sin B =n then, prove that (m + n°) cos' B=n
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hi frnd
M = cos Acos B; n = cos Asin Bnow, LHS = (m2+n2) cos2B=[cos2Acos2B + cos2Asin2B] . cos2B=[cos2A .sin2B + cos2A . cos2Bcos2B . sin2B] . cos2B=cos2A(sin2B+ cos2B)sin2B=cos2Asin2B=n2=RHS
hope this helps you
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