Question

Prove the identity
(2tanx)/(x+tan^2x) = sin2x

Answer 1

@freckles haha sorry me again..with more trig.
I've gotten
[(2tanx)/(1-tan^2x)][cos^2x]

Answer 2

Ah no, that's wrong. Let me try again xD

Answer 3

1) (2tanx)/(sec^2x)

Answer 4

Gasp!! I think I got it :)
\[\frac{ 2tanx }{ 1+\tan^2x }\] \[\frac{ 2*\frac{ sinx }{ cosx } }{ \frac{ 1 }{ \cos^2x } }\] \[2*\frac{ sinx }{ cosx }*\frac{ \cos^2x }{ 1 }\]
= simplify and cross out a few things
2sinxcosx

Answer 5

@ganeshie8 is this correct? :)

Answer 6

@jim_thompson5910

Answer 7

then you'd turn 2*sin(x)*cos(x) into sin(2x)

Answer 8

make sure to keep the right hand side the same the whole time

Answer 9

after you reach
2sinxcosx

use the identity
sin2x = 2sinxcosx and you'll have
sin2x = sin 2x

Answer 10

wait... is there a typo somewhere?
you had
\[\frac{2tanx}{x+\tan^2x}\]
in your original question and then all of a sudden I see

\[\frac{2tanx}{1+\tan^2x}\]
I am crossing fingers here because if that x+tan^2x turns out to be a typo then everything will work out

Answer 11

Right right, I just wanted to make sure the process was correct :)

Answer 12

um..the second one hahaa sorry yeah typo.

Answer 13

\[\frac{ 2tanx }{ 1+\tan^2x } = \sin2x \]


\[\frac{ 2tanx }{ \sec^2x } =\sin 2x \]
\[\frac{ 2\frac{sinx}{cosx} }{ \frac{1}{\cos^2x}} = sin2x \]
\[2\frac{sinx}{cosx} \times \frac{\cos^2x}{1} = \sin 2x \]
the cos x cancels out and 2/1 = 2
\[2sinxcosx = 2sinx \]
using identity
2sinxcosx = sin2x
\[\sin2x = \sin2x \]

Answer 14

your attempt is correct...just have to remember that there are A LOT of trig. identities.

Answer 15

Right...So I had it all right :)
I just didn't do the last step to make it into sin2x.

Answer 16

thanks so much!!

Answer 17

very great work @Babynini

Answer 18

thank you :)