Question

Solution of the equation $\frac{1+\tan x+{\tan }^{2}x+....+ta{n}^{n}x+...}{1-\tan x+ta{n}^{2}x-...+{(-1)}^n{\tan }^{n}x+...}=1+\sin 2x$ is

• A. $n\pi$
• B. $2n\pi$
• C. $(2n\pm 1)\frac{\pi }{2}$
• D. None of these

Answer 1

Answer: (A)

$n\pi$

$\Rightarrow$$\frac{1+\tan x+{\tan }^{2}x+....+ta{n}^{n}x+...}{1-\tan x+ta{n}^{2}x-...+{(-1)}^n{\tan }^{n}x+...}=1+\sin 2x$

$\Rightarrow$ $S_{\infty }=\frac{a}{1-r}$........Formula

$\Rightarrow$$\frac{\frac{1}{1-\tan x}}{\frac{1}{1+\tan x}}=1+\frac{2\tan x}{1+{\tan }^{2}x}$

$\Rightarrow$$\frac{1+\tan x}{1-\tan x}=\frac{1+{\tan }^{2}x+2\tan x}{1+{\tan }^{2}x}$

$\Rightarrow {\tan }^{3}x+{\tan }^{2}x=0$

$\Rightarrow \tan x=0,{\tan }^{2}x=-1$ (not possible)

$\Rightarrow x=n\pi$