Question

If $\tan x+2\tan 2x+3\tan 3x+8\cot 8x=\sqrt{3}$ then the general solution of $x=$

• A. $n\pi +\frac{\pi }{3},\forall \in Z$
• B. $n\pi +\frac{\pi }{6},\forall \in Z$
• C. $n\pi +\frac{\pi }{4},\forall \in Z$
• D. $n\pi ,\forall \in Z$

Answer 1

The correct option is B $n\pi +\frac{\pi }{6},\forall \in Z$

given,$tanx+2tan2x+4tan4x+8cot8x=\sqrt{3}$

To solve the expression . we first find tanx -cotx

$\to tanx-cotx$

$\to tnax-1/tanx$

$\to \frac{ta{n}^2-1}{tanx}$

Multiplaying 2 in both numerator & denominator

$\to -2\frac{(1-ta{n}^{2}x)}{2tanx}\to \frac{-2}{\frac{2tanx}{1-ta{n}^{2}x}}\to \frac{-2}{tan2x}\to -2cot2x$

So, $tanx-cotx=-2cot2x$

Now, let $a=tanx+2tan2x+4tan4x+8cot8x$

Add & subtract cotx in a

$\to a=(tanx-cotx)+2tan2x+4tan4x+8cot8x+cot8x$

$\to a=-2cot2x+2tan2x+4tan4x+8cot8x+cotx$

$\to a=+2(tan2x-cotx)+4tan4x+8cot8x+cotx$

$\to a=-4cot4x+4tan4x+8cot8x+cotx$

$\to a=4(tan4x-cot4x)+8cot8x+cotx$

$\to a=-8cot8x+8cot8x+cotx$

$\to a=cotx$

$\therefore cotx=\sqrt{3}$

The solution is $n\pi +\frac{\pi }{6},\forall z$