Proving Trig identities (1 Viewer)

[kazemagic]

How to prove the above ^^ Thanks!

 

[Nooblet94]

How to prove the above ^^ Thanks!

<a href="http://www.codecogs.com/eqnedit.php?latex=\begin{align*} LHS &=\frac{1@plus;\sin A}{1-\sin A}\\ &=\frac{(1@plus;\sin A)^2}{1-\sin^2A}\\ &=\frac{(1@plus;\sin A)^2}{\cos^2A}\\ &=\left (\frac{1@plus;\sin A}{\cos A}\right )^2\\ &=\left (\frac{1}{\cos A}@plus;\frac{\sin A}{\cos A} \right )^2\\ &=\left (\frac{1}{\cos A}@plus;\tan A \right)^2\\ &=RHS \end{align*}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\begin{align*} LHS &=\frac{1+\sin A}{1-\sin A}\\ &=\frac{(1+\sin A)^2}{1-\sin^2A}\\ &=\frac{(1+\sin A)^2}{\cos^2A}\\ &=\left (\frac{1+\sin A}{\cos A}\right )^2\\ &=\left (\frac{1}{\cos A}+\frac{\sin A}{\cos A} \right )^2\\ &=\left (\frac{1}{\cos A}+\tan A \right)^2\\ &=RHS \end{align*}" title="\begin{align*} LHS &=\frac{1+\sin A}{1-\sin A}\\ &=\frac{(1+\sin A)^2}{1-\sin^2A}\\ &=\frac{(1+\sin A)^2}{\cos^2A}\\ &=\left (\frac{1+\sin A}{\cos A}\right )^2\\ &=\left (\frac{1}{\cos A}+\frac{\sin A}{\cos A} \right )^2\\ &=\left (\frac{1}{\cos A}+\tan A \right)^2\\ &=RHS \end{align*}" /></a>
Also, the quickest way to post an equation from CodeCogs is to go to the box at the bottom where it has the URL and select HTML (Edit). Then you can just copy and paste that code into your post.

 

[Ferox]

Multiply the LHS by (1+sinA)/(1+sinA). The top is now (1+sinA)^2 and the bottom is the difference of two squares 1-(sinA)^2 = (cosA)^2
LHS = (1+sinA)^2/(cosA)^2 = [(1+sinA)/cosA]^2 = [1/cosA + tanA]^2

 

[kazemagic]

I got to the 4th last line and didn't know I had to factorise the power of 2(face palm lol)
Yes, I will select HTML in the future, first time using a maths equation editor lol
Thanks for the answer!

 

[kazemagic]

Got another question,
<a href="http://www.codecogs.com/eqnedit.php?latex=(cosA@plus;\frac{1}{cosA})\times\frac{1}{tanA}=\frac{cos^2A@plus;1}{sinA}" target="_blank"><img src="http://latex.codecogs.com/png.latex?(cosA+\frac{1}{cosA})\times\frac{1}{tanA}=\frac{cos^2A+1}{sinA}" title="(cosA+\frac{1}{cosA})\times\frac{1}{tanA}=\frac{cos^2A+1}{sinA}" /></a>

Thanks!

 

[nightweaver066]

 

[zeebobDD]

make a common factor for cosa + 1/cosa then multiply it by 1/tana knowing that tan a is equal to sina/cosa and theres your answer

 

[kazemagic]

ahh, thanks guys!