Question

1/tanx + tanx=sec^2x/tanx

Answer 1

1/tanx + tanx=sec^2x/tanx

SAY: tanx= tan^2x/tanx

this way you put everything under a denominator of tanx which would mean that all denominators of tanx cancel.

Answer 2

1/tanx + tanx = sec^2x/tanx

1/tanx + tan^2x/tanx = sec^2x/tanx multiply everything by tanx,

1 + tan^2x = sec^2x

identity proved.

Answer 3

This only works if you were allowed to manipulate both sides, but if you were only able to work the one side, work the LEFT SIDE.

1/tanx + tanx = sec^2x/tanx

1/tanx + tan^2x/tanx = sec^2x/tanx add the fractions on the left

1 + tan^ 2x /tanx = sec^2x/tanx

1+tan^2x=sec^2x, so the left side becomes sec^2x/tanx .

Hence is the proof, result is,

sec^2x/tanx =sec^2x/tanx

Answer 4

I dont understand how 1+tan^2x/tanx could equal 1+tan^2x, wouldn't the tan^2x being divided by tanx make it equal 1+tanx?

Answer 5

Lets start over, okay?

Answer 6

Tan x = Tan ^2x / Tanx right?

Answer 7

yes

Answer 8

so, I am substituting Tan ^2x / Tanx instead of Tan x.

\[\frac{1}{Tanx}+\color{red}{Tanx }=\frac{Sec^2x}{Tanx} \]

\[\frac{1}{Tanx}+\color{red}{ \frac{Tan^2x}{Tanx}} =\frac{Sec^2x}{Tanx} \]

Now, I am adding fractions on the left side,
\[\frac{1+Tan^2x}{Tanx} =\frac{Sec^2x}{Tanx} \]

Answer 9

Knowing that \[\color{red}{ Tan^2x+1=Sec^2x}\] your identity is proved with a last substitution

(substitute Sec^2x for 1+Tan^2x )

Answer 10

OH! Wow okay, so obvious. I don't know why it I couldn't get it through my brain. Thank you!

Answer 11

Becuase I didn't post it in latex. This latex just gives your eyes the chance to see it :)

Answer 12

YW!

Answer 13

would you mind helping me with one more?

Answer 14

Sure.

Answer 15

It's probably simple too, but it is csc(-x)/sec(-x)=-cot
I got to -sin/cosx=-cot but Im not sure how to make what I have into the cotangent when it is just tangent right now

Answer 16

\[\huge\color{blue}{ \frac{\csc(-x)}{\sec(-x)} =-\cot(x) } \] this?

Answer 17

yes