Prove that tan 4 x=frac{4 tan xleft(1-tan ^{2 | vedclass

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Question

It is known that $	an 2 A=\frac{2 	an A}{1-	an ^{2} A}$

$	herefore$ $L.H.S.$ $=	an 4 x=	an 2(2 x)$

$=\frac{2 	an 2 x}{1-	an ^{2}(2 x)}$

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Vedclass · Answer

It is known that $	an 2 A=\frac{2 	an A}{1-	an ^{2} A}$

$	herefore$ $L.H.S.$ $=	an 4 x=	an 2(2 x)$

$=\frac{2 	an 2 x}{1-	an ^{2}(2 x)}$

$=\frac{2\left(\frac{2 	an x}{1-	an ^{2} x}ight)}{1-\left(\frac{2 	an x}{1-	an ^{2} x}ight)^{2}}$

$=\frac{\left(\frac{4 	an x}{1-	an ^{2} x}ight)}{\left[1-\frac{4 	an ^{2} x}{\left(1-	an ^{2} xight)^{2}}ight]}$

$=\frac{\left(\frac{4 	an x}{1-	an ^{2} x}ight)}{\left[\frac{\left(1-	an ^{2} xight)^{2}-4 	an ^{2} x}{\left(1-	an ^{2} xight)^{2}}ight]}$

$=\frac{4 	an x\left(1-	an ^{2} xight)}{\left(1-	an ^{2} xight)^{2}-4 	an ^{2} x}$

$=\frac{4 	an x\left(1-	an ^{2} xight)}{1+	an ^{4} x-2 	an ^{2} x-4 	an ^{2} x}$

$=\frac{4 	an x\left(1-	an ^{2} xight)}{1-6 	an ^{2} x+	an ^{4} x}= R . H.S.$

Prove that $\tan 4 x=\frac{4 \tan x\left(1-\tan ^{2} x\right)}{1-6 \tan ^{2} x+\tan ^{4} x}$

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