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

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

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

Vedclass · Accepted Answer

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

Similar Questions

Prove that $\sin ^{2} 6 x-\sin ^{2} 4 x=\sin 2 x \sin 10 x$

$2{\cos ^2}\theta - 2{\sin ^2}\theta = 1$, then $\theta =$ .......$^o$

If $\tan \alpha = \frac{1}{7},\;\tan \beta = \frac{1}{3},$ then $\cos 2\alpha = $

$2\cos x - \cos 3x - \cos 5x = $

$\sin 4\theta $ can be written as