# Math Help - Quick one about trig identities

1. ## Quick one about trig identities

Hey there MHF.

I got a quick question with hopefully a quick answer.

How does $\frac{sin^2x}{cos^2x}+1=sec^2x$ ?

Thanks!

2. ## Re: Quick one about trig identities

Originally Posted by Paze
Hey there MHF.

I got a quick question with hopefully a quick answer.

How does $\frac{sin^2x}{cos^2x}+1=sec^2x$ ?

Thanks!
\displaystyle \begin{align*} \sin^2{(x)} + \cos^2{(x)} &= 1 \\ \frac{\sin^2{(x)} + \cos^2{(x)}}{\cos^2{(x)}} &= \frac{1}{\cos^2{(x)}} \\ \frac{\sin^2{(x)}}{\cos^2{(x)}} + 1 &= \sec^2{(x)} \end{align*}

3. ## Re: Quick one about trig identities

it's simple, look:

to get the proof of this identity you have to choose a side of equation to start and get the other side. Using the left side we have:

$\dfrac{\sin^2 x}{\cos^2 x}+1=\dfrac{\sin^2 x +\cos^2 x}{\cos^2 x}=\dfrac{1}{\cos^2 x}=\sec^2 x$

I hope you understood, greetings

4. ## Re: Quick one about trig identities

Thanks guys. I did not realize that sin(x)+cos(x)=1. Is there a quick proof for that one or? Is it only when they are squared maybe..?

5. ## Re: Quick one about trig identities

No! be carefully it's $cos^2 x + \sin^2 x =1$ pay attention with the squares! that's called "fundamental trigonometric's identity" and it's valid for all $x \in \mathbb{R}$

there's a proof I think you can find it in wiki or any precalculus book

6. ## Re: Quick one about trig identities

Think of the unit circle. If you draw in a radius, it makes an angle \displaystyle \begin{align*} \theta \end{align*} with the positive x-axis. Then draw a segment from the end of the radius to the x-axis, you will have a right-angle triangle. The horizontal length is \displaystyle \begin{align*} \cos{(\theta)} \end{align*} and the vertical length is \displaystyle \begin{align*} \sin{(\theta)} \end{align*}. Since it's a right-angle-triangle, Pythagoras' Theorem holds, and so \displaystyle \begin{align*} \left[ \sin{(\theta)} \right] ^2 + \left[ \cos^2{(\theta)} \right]^2 = 1^2 \end{align*}, or \displaystyle \begin{align*} \sin^2{(\theta)} + \cos^2{(\theta)} = 1 \end{align*}. This is true no matter the value of \displaystyle \begin{align*} \theta \end{align*}.

7. ## Re: Quick one about trig identities

Originally Posted by Prove It
Think of the unit circle. If you draw in a radius, it makes an angle \displaystyle \begin{align*} \theta \end{align*} with the positive x-axis. Then draw a segment from the end of the radius to the x-axis, you will have a right-angle triangle. The horizontal length is \displaystyle \begin{align*} \cos{(\theta)} \end{align*} and the vertical length is \displaystyle \begin{align*} \sin{(\theta)} \end{align*}. Since it's a right-angle-triangle, Pythagoras' Theorem holds, and so \displaystyle \begin{align*} \left[ \sin{(\theta)} \right] ^2 + \left[ \cos^2{(\theta)} \right]^2 = 1^2 \end{align*}, or \displaystyle \begin{align*} \sin^2{(\theta)} + \cos^2{(\theta)} = 1 \end{align*}. This is true no matter the value of \displaystyle \begin{align*} \theta \end{align*}.
Spot on as always! Thank you, Prove It!