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if α and β are acute angles and cos2α=3cos2β-1/3-cos2β , show that tanα=√2 tanβ
7 years ago

Answers : (1)

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Jitender Singh
IIT Delhi
askIITians Faculty
158 Points 
							Ans:Hello student, please find answer to your question
tan\alpha = \sqrt{2}tan\beta
cos2\alpha = \frac{1-tan^{2}\alpha }{1+tan^{2}\alpha }
RHS = cos2\alpha = \frac{1-(\sqrt{2}tan\beta )^{2} }{1+(\sqrt{2}tan\beta )^{2}}
RHS = cos2\alpha = \frac{1-2tan^{2}\beta }{1+2tan^{2}\beta }
LHS = \frac{3cos2\beta -1}{3-cos2\beta }
LHS = \frac{3.\frac{1-tan^{2}\beta }{1+tan^{2}\beta } -1}{3-\frac{1-tan^{2}\beta }{1+tan^{2}\beta } }
LHS = \frac{3-3tan^{2}\beta-1-tan^{2}\beta }{3+3tan^{2}\beta-1+tan^{2}\beta }
LHS = \frac{2-4tan^{2}\beta}{2+4tan^{2}\beta}
LHS = \frac{1-2tan^{2}\beta}{1+2tan^{2}\beta}
Hence, proved
5 years ago
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