Question

$if \: tan \: a = cos2 \alpha \: then \: prove \: that \: \\ sin2a = \frac{1 - {tan}^{4} \alpha }{1 + {tan}^{4} \alpha }$
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Answer 1

Answer:

Given,

\frac{\sec8\alpha\;-\;1}{\sec4\alpha\;-\;1}\\\;\\=\frac{\frac{1}{\cos8\alpha}-1}{\frac{1}{\cos4\alpha}-1}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{1-\cos8\alpha}{1-\cos4\alpha}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{1- 1 +2\sin^{2}4\alpha}{1- 1 +2\sin^{2}2\alpha}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{2\sin^{2}4\alpha}{2\sin^{2}2\alpha}\\\;\\=\frac{\cos4\alpha}{\cos8\alpha}.\frac{\sin^{2}4\alpha}{\sin^{2}2\alpha}\\\;\\=\frac{2\sin4\alpha.\cos4\alpha.2\sin2\alpha.\cos2\alpha}{2\cos8\alpha.\sin^{2}2\alpha}\\\;\\=\frac{\sin8\alpha.\cos2\alpha}{\cos8\alpha.sin2\alpha}\\\;\\=\frac{tan8\alpha}{tan2\alpha}\\\;\\Note:1. cos2\theta= 1-2\sin^{2}\theta\\\;\\\sin2A=2\sin A.\cos A