Trigonometry> trigonometry...

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

hannah mannah , 13 Years ago

Grade

1 Answers

Jitender Singh

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

Last Activity: 11 Years ago