Question

If (2tan(x))/(1-tan^2(x))=1, then what can x equal?
My options are:
A. x=7pi/8+npi
B. x=pi/8+npi
C. x=3pi/8+npi
D. x=5pi/8+npi

Answer 1

I think that B is right but I don't know about any others. Can anybody help me?

Answer 2

I got the question wrong, but I still want to know how to do this!

Answer 3

\[\frac{ 2\tan x }{ 1-\tan ^2x }=1\]
\[\tan 2x=1=\tan \left( \frac{ \pi }{ 4 }+n \pi \right)\]
\[x=\left( \frac{ \pi }{ 8 }+n \frac{ \pi }{ 2 } \right)\]

Answer 4

How did you go between the last two steps?

Answer 5

\[\tan 2 x=\frac{ \sin 2x }{ \cos 2x }=\frac{ 2\sin x \cos x }{ \cos ^2x-\sin ^2x }\]
divide the numerator and denominator by \[\cos ^2x\]
\[\tan 2x=\frac{ 2\tan x }{ 1-\tan ^2x }\]

Answer 6

\[\tan 2x=1=\tan \frac{ \pi }{ 4 }=\tan \left( \frac{ \pi }{ 4 }+n \pi \right)\]
\[2x=\frac{ \pi }{ 4 }+n \pi,x=?\]

Answer 7

So that's \[1/8\pi +1/2n \pi ?\]

Answer 8

correct ,i have written above.

Answer 9

Okay I get it thank you. That helps a lot.

Answer 10

You can have a medal :)

Answer 11

yw

Answer 12

can you help me with another problem? @surjithayer

Answer 13

I have \[\cos(\pi/4)\cos(\pi/6)=1/2( (blank) \pi/12+\cos(5\pi/12)\]

Answer 14

\[2 \cos A \cos B=\cos \left( A+B \right)+\cos \left( A-B \right)\]
\[\cos A \cos B=\frac{ 1 }{ 2 }\left\{ \cos \left( A+B \right)+\cos \left( A-B \right) \right\}\]