# Math Help - Verify the trigonometric identity

1. ## Verify the trigonometric identity

I'm currently learning trigonometric identities and my favourite way of verifying a given trigonometric identity is to transform one side to the other.
I usually do this by reducing the terms of the side am working on down to sines and cosines and then manipulating them with algebra and the
basic trigonometric identities. I love this method because it makes it so clear to see how both sides are equivalent that even if I can prove an
identity some other way I still use this.
But am having trouble using this method on trigonometric identities expressed as fractions with more than 1 term in the denominator.
For example, could someone show me how one would go about employing this method to verify the following by transforming the left hand side to the right hand side?

$\frac{tan\: \beta}{1 - cot\: \beta} + \frac{cot\: \beta}{1 - tan\: \beta} = tan\: \beta + cot\: \beta + 1$

2. ## Re: Verify the trigonometric identity

Originally Posted by kodx
I'm currently learning trigonometric identities and my favourite way of verifying a given trigonometric identity is to transform one side to the other.
I usually do this by reducing the terms of the side am working on down to sines and cosines and then manipulating them with algebra and the
basic trigonometric identities. I love this method because it makes it so clear to see how both sides are equivalent that even if I can prove an
identity some other way I still use this.
But am having trouble using this method on trigonometric identities expressed as fractions with more than 1 term in the denominator.
For example, could someone show me how one would go about employing this method to verify the following by transforming the left hand side to the right hand side?

$\frac{tan\: \beta}{1 - cot\: \beta} + \frac{cot\: \beta}{1 - tan\: \beta} = tan\: \beta + cot\: \beta + 1$
\displaystyle \begin{align*} \frac{\tan{(\beta)} }{1 - \cot{(\beta)}} + \frac{\cot{(\beta)}}{1 - \tan{(\beta)}} &= \frac{\frac{\sin{(\beta)}}{\cos{(\beta)}}}{1 - \frac{\cos{(\beta)}}{\sin{(\beta)}}} + \frac{\frac{\cos{(\beta)}}{\sin{(\beta)}}}{1 - \frac{\sin{(\beta)}}{\cos{(\beta)}}} \\ &= \frac{\frac{\sin{(\beta)}}{\cos{(\beta)}}}{\frac{ \sin{(\beta)} - \cos{(\beta)}}{\sin{(\beta)}}} + \frac{\frac{\cos{(\beta)}}{\sin{(\beta)}}}{\frac{ \cos{(\beta)} - \sin{(\beta)}}{\cos{(\beta)}}} \\ &= \frac{\sin^2{(\beta)}}{\cos{(\beta)} \left[ \sin{(\beta)} - \cos{(\beta)} \right] } + \frac{\cos^2{(\beta)}}{\sin{(\beta)} \left[ \cos{(\beta)} - \sin{(\beta)} \right] } \\ &= \frac{\sin^2{(\beta)}}{\cos{(\beta)} \left[ \sin{(\beta)} - \cos{(\beta)} \right] } - \frac{\cos^2{(\beta)}}{\sin{(\beta)} \left[ \sin{(\beta)} - \cos{(\beta)} \right] } \end{align*}

\displaystyle \begin{align*} &= \frac{\sin^3{(\beta)}}{ \sin{(\beta)} \cos{(\beta)} \left[ \sin{(\beta)} - \cos{(\beta)} \right] } - \frac{\cos^3{(\beta)}}{ \sin{(\beta)} \cos{(\beta)} \left[ \sin{(\beta)} - \cos{(\beta)} \right] } \\ &= \frac{\sin^3{(\beta)} - \cos^3{(\beta)} }{ \sin{(\beta)} \cos{(\beta)} \left[ \sin{(\beta)} - \cos{(\beta)} \right] } \\ &= \frac{ \left[ \sin{(\beta)} - \cos{(\beta)} \right] \left[ \sin^2{(\beta)} + \sin{(\beta)} \cos{(\beta)} + \cos^2{(\beta)} \right] }{ \sin{(\beta)} \cos{(\beta)} \left[ \sin{(\beta)} - \cos{(\beta)} \right] } \\ &= \frac{\sin^2{(\beta)} + \sin{(\beta)}\cos{(\beta)} + \cos^2{(\beta)}}{\sin{(\beta)}\cos{(\beta)}} \\ &= \frac{ \sin^2{(\beta)} }{ \sin{(\beta)} \cos{(\beta)} } + \frac{ \sin{(\beta)} \cos{(\beta)} }{ \sin{(\beta)} \cos{(\beta)} } + \frac{ \cos^2{(\beta)} }{ \sin{(\beta)} \cos{(\beta)} } \\ &= \tan{(\beta)} + 1 + \cot{(\beta)} \end{align*}

3. ## Re: Verify the trigonometric identity

Or you can use this approach:

$\frac {\tan \beta}{1- cot \beta} + \frac {cot \beta}{1-\tan \beta} = \frac {\tan \beta(1- \tan \beta) + cot \beta(1- cot \beta)}{(1- cot \beta)(1-\tan \beta)}$

$= \frac {\tan \beta - \tan^2 \beta + \frac 1 {\tan \beta} - \frac 1 {\tan^2 \beta}}{2 - \frac 1 {tan \beta} - \tan \beta}$

Multiply top and bottom by $\tan^2 \beta$:

$= \frac {- \tan^4 \beta + \tan ^3 \beta+ \tan \beta -1}{-\tan^3 \beta + 2 \tan^2 \beta - \tan \beta}$

and factor the top:

$= \frac {(\tan^2 \beta + \tan \beta + 1)(-\tan^2 \beta+ 2\tan \beta -1)}{\tan \beta (-\tan^2 \beta+ 2\tan \beta -1)}$

$= \tan \beta + 1 + \frac 1 {\tan \beta}$

Phew!

4. ## Re: Verify the trigonometric identity

Hello, kodx!

$\frac{\tan x}{1 - \cot x} + \frac{\cot x}{1 - \tan x} \:=\:\tan x + \cot x + 1$

We have: . $\dfrac{\tan x}{1 - \frac{1}{\tan x}} + \frac{\frac{1}{\tan x}}{1 - \tan x}$

Multiply both fractions by $\tfrac{\tan x}{\tan x}$

. . $\dfrac{\tan x}{1-\frac{1}{\tan x}}\cdot{\color{blue}\frac{\tan x}{\tan x}} + \frac{\frac{1}{\tan x}}{1-\tan x}\cdot{\color{blue}\frac{\tan x}{\tan x}} \;=\;\frac{\tan^2\!x}{\tan x - 1} + \frac{1}{\tan x(1-\tan x)}$

. . $=\;\frac{\tan^2\!x}{\tan x-1} - \frac{1}{\tan x(\tan x - 1)} \;=\; \frac{\tan^3\!x}{\tan x(\tan x-1)} - \frac{1}{\tan x(\tan x -1)}$

. . $=\;\frac{\tan^3\!x-1}{\tan x(\tan x-1)} \;=\;\frac{(\tan x - 1)(\tan^2\!x + \tan x + 1)}{\tan x(\tan x-1)}$

. . $=\;\frac{\tan^2\!x + \tan x + 1}{\tan x} \;=\;\frac{\tan^2\!x}{\tan x} + \frac{\tan x}{\tan x} + \frac{1}{\tan x}$

. . $=\;\tan x + 1 + \cot x$

5. ## Re: Verify the trigonometric identity

Thanks Prove It. Your explanation's really well detailed. That's actually how I tried it the first time but I was afraid of getting cubes so I stopped short at introducing more sines or cosines. I guess I should stop being so scared of dealing with cubes.

6. ## Re: Verify the trigonometric identity

Phew!
Phew! indeed. But that's actually a lot more succinct than what I had in mind. I guess I don't always have to reduce everything down to sines and cosines. But like I said above, I think I've just realized that my real problem was being afraid to introduce expressions with powers greater than 2 afraid I wouldn't be able to factor them out. Turns out that was exactly the way to go.
So am going to work on my attitude towards cubic (and from your reply, quartic) expressions. Maybe I'll find out they're just as scared of me as I am of them.