If n=CosA/CosB m=SinA/SinB So prove that- (m2-n2)Sin^2B=1-n 2

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Aditya Gupta · Answer

wehave to prove (m2-n2)Sin^2B=1-n 2 or equivalently m^2sin^B+n^2cos^2B=1 given that n=CosA/CosB and m=SinA/SinB so, LHS = sin^2A+cos^2A=1=RHS hence proved

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If n=CosA/CosB m=SinA/SinB So prove that- (m2-n2)Sin^2B=1-n 2

If n=CosA/CosB
m=SinA/SinB
So prove that-
^{(m2-n2)Sin^2B=1-n}²

1 Answers

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If n=CosA/CosB m=SinA/SinB So prove that- (m2-n2)Sin^2B=1-n 2

If n=CosA/CosB m=SinA/SinBSo prove that- (m2-n2)Sin^2B=1-n2

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If n=CosA/CosB
m=SinA/SinB
So prove that-
^{(m2-n2)Sin^2B=1-n}²