Question

Identity transformations of Trigonometric Expressions.
prove the following identities.
$\frac{si{n}^{2}2a+4si{n}^4\alpha -4si{n}^2\alpha co{s}^2\alpha }{4-si{n}^{2}2\alpha -4si{n}^2\alpha }=ta{n}^4\alpha .$

Answer 1

$\frac{si{n}^{2}2\alpha +4si{n}^4\alpha -4si{n}^{2}co{s}^2\alpha }{4-si{n}^{2}2\alpha -4si{n}^2\alpha }=LHS$

$sin2\alpha =2sin\alpha cos\alpha$

$=\frac{si{n}^{2}2\alpha +4si{n}^4\alpha -si{n}^{2}2\alpha }{4(1-si{n}^2\alpha )-si{n}^{2}2\alpha }$

$=\frac{4si{n}^4\alpha }{4co{s}^2\alpha -si{n}^{2}2\alpha }$ $\because si{n}^2\alpha +co{s}^2\alpha =1$

$=\frac{4si{n}^2\alpha }{4co{s}^2\alpha -4si{n}^2\alpha co{s}^2\alpha }$

$=\frac{4si{n}^4\alpha }{4(co{s}^2\alpha -si{n}^2\alpha co{s}^2\alpha )}$

$=\frac{si{n}^4\alpha }{co{s}^2\alpha (1-si{n}^2\alpha )}$

$=ta{n}^4\alpha$

$=RHS$

Hence proved!