1+tan^2 x/tan^2 x = csc^2 X
I tried to prove this identity and I couldn't plz help Thanks...
Hello,
Transform $\displaystyle \tan^2x=\frac{\sin^2x}{\cos^2x}$
-->
$\displaystyle \begin{aligned} \frac{1+tan^2x}{tan^2x} &=\frac{1+\frac{\sin^2x}{\cos^2x}}{\frac{\sin^2x}{ \cos^2x}} \\
&=\left(1+\frac{\sin^2x}{\cos^2x}\right) \cdot \frac{\cos^2x}{\sin^2x} \\
&=\left(\cos^2x+\sin^2x\right) \cdot \frac{1}{\sin^2x} \\
&=\dots \end{aligned}$
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
Hello, victorfk06!
We're expected to know some basic identities . . .
Prove: .$\displaystyle \frac{1+\tan^2\!x}{\tan^2\!x}\:= \:\csc^2\!x$
$\displaystyle \text{We have: }\;\overbrace{\frac{1}{\tan^2\!x}}^{\text{This is }\cot^2\!x} + \frac{\tan^2\!x}{\tan^2\!x} \;\;=\;\;\underbrace{\cot^2\!x + 1}_{\text{This is }\csc^2\!x} \;\;=\;\;\csc^2\!x$