Question

tan x + 1 = sqrt3 + sqrt3 cot x

Answer 1

factor out of sqrt3 on right side, can you do that?

Answer 2

\[\tan x + 1 = \sqrt{3} + \sqrt{3} \cot x\]\[\tan x + 1 = (1+\cot x)\sqrt{3} \]\[\frac{1 + \tan x}{1+\cot x}=\sqrt{3}\]\[\frac{(1 + \tan x)(1-\cot x)}{(1+\cot x)(1-\cot x)}=\sqrt{3}\]

Answer 3

can i have help with another?

Answer 4

\[\frac{1 + \tan x -\cot x+1}{1-\cot^2x}=\sqrt{3}\]

Answer 5

\[\large \frac{2 + \tan x - \frac{1}{\tan x}}{1-\cot^2x}=\sqrt{3}\]\[\large \frac{\frac{2\tan x}{\tan x} + \frac{\tan^2x}{\tan x} - \frac{1}{\tan x}}{1-\cot^2x}=\sqrt{3}\]\[\Large \frac{\frac{2\tan x+\tan^2x-1}{\tan x} }{1-\cot^2x}=\sqrt{3}\]\[\Large \frac{\frac{2\tan x+\tan^2x-1}{\tan x} }{1-\frac{1}{\tan^2x}}=\sqrt{3}\]\[\Large \frac{\frac{2\tan x+\tan^2x-1}{\tan x} }{\frac{\tan^2x}{\tan^2x}-\frac{1}{\tan^2x}}=\sqrt{3}\]\[\Large \frac{\frac{2\tan x+\tan^2x-1}{\tan x} }{\frac{\tan^2x-1}{\tan^2x}}=\sqrt{3}\]

Answer 6

\[\Large \frac{(2\tan x+\tan^2x-1)\tan x}{\tan^2x-1} =\sqrt{3}\]

Answer 7

\[\Large \frac{(\tan^2x+2\tan x-1)\tan x}{\tan^2x-1} =\sqrt{3}\]

Answer 8

Thank you! I tried to look at this before, but my computer wouldn't load. Sorry!