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Question

If cos Bsin A=n and cos Bcos A=m, then show that (m2+n2)cos2A=n2.

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Solution

n=cos Bsin A;m=cos Bcos A
So, n2=cos2Bsin2A; m2=cos2Bcos2A
L.H.S.=(m2+n2)cos2A=(cos2Bcos2A+cos2Bsin2A)cos2A=sin2A cos2B+cos2A cos2Bcos2A sin2A×cos2A=cos2B(sin2A+cos2A)sin2A=cos2Bsin2A
=n2=R.H.S. Hence Proved.

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