1. ## Prove the IDENTITY

1+tan^2 x/tan^2 x = csc^2 X

I tried to prove this identity and I couldn't plz help Thanks...

2. Hello,

Originally Posted by victorfk06
1+tan^2x/tan^2x = csc^2 X

I tried to prove this identity and I couldn't plz help Thanks...
Transform $\tan^2x=\frac{\sin^2x}{\cos^2x}$

-->
\begin{aligned} \frac{1+tan^2x}{tan^2x} &=\frac{1+\frac{\sin^2x}{\cos^2x}}{\frac{\sin^2x}{ \cos^2x}} \\
&=\left(1+\frac{\sin^2x}{\cos^2x}\right) \cdot \frac{\cos^2x}{\sin^2x} \\
&=\left(\cos^2x+\sin^2x\right) \cdot \frac{1}{\sin^2x} \\
&=\dots \end{aligned}

3. Then what I didn't understand it?

4. Originally Posted by victorfk06
Then what I didn't understand it?
Erm... pardon me , but are you saying you wonder why you didn't see it, or that you still don't understand it ?

Here's a tip, for the future :

You're with a tan function, and you want to get to sines (csc is just 1/sin). So one of the ways (it depends on the situation) is to remember that the relation between tan and sin is :

tan=sin/cos

5. Oh thanks dats the final answer correct?

6. I'm confused lol !

What follows my expressions is :

\begin{aligned} \dots &=(\sin^2x+\cos^2x) \cdot \frac{1}{\sin^2x} \\
&=1 \cdot \frac{1}{\sin^2x} \\
&=\left(\frac{1}{\sin x}\right)^2 \\
&=csc^2 x \end{aligned}

7. Hello, victorfk06!

We're expected to know some basic identities . . .

Prove: . $\frac{1+\tan^2\!x}{\tan^2\!x}\:= \:\csc^2\!x$

$\text{We have: }\;\overbrace{\frac{1}{\tan^2\!x}}^{\text{This is }\cot^2\!x} + \frac{\tan^2\!x}{\tan^2\!x} \;\;=\;\;\underbrace{\cot^2\!x + 1}_{\text{This is }\csc^2\!x} \;\;=\;\;\csc^2\!x$