Proving this trig equation

[jwpaine]

I have the equation:

\(\displaystyle \L (cotx\,+\,cscx)^{2} = \frac{secx\,+\,1}{secx\,-1}\)

I know the reciprocal identities, Pythagorean identities and quotient identities...which should be sufficient for proving this equation.

I started to expand the left hand side to

\(\displaystyle \L cot^{2}x\,+\,2cotx \cdot cscx\,+\,csc^{2}x\)

What would be my best, next step?


John.

 

[tkhunny]

Your question at the end tells the story of greatest importance. You are looking for the wrong thing! Just find A step - one that will or will not help you. Just try something. Looking for the best way is a pursuit for impressing friends and boring relatives. Just take a path and see where it goes. You will not break it.

Personally, it looks to me like you have made it worse.

Rather than doing that, what happens if you multiply the righthand side by \(\displaystyle \frac{\cos(x)}{\cos(x)}\), causing it to look like \(\displaystyle \frac{1+\cos(x)}{1-\cos(x)}\). I'm already starting to be suspicious about the square on the left and no square on the right. This leads me to consider multiplying the righthand side by \(\displaystyle \frac{1+\cos(x)}{1+\cos(x)}\), giving \(\displaystyle \frac{(1+\cos(x))^{2}}{\sin^{2}(x)}\). This is tantalizingly close to where we want to be. My last suspiscion is the sine and cosine on the righthand side, but no sine or cosine on the left. That's easy enough to fix.

 

[jwpaine]

I decided to go for the left hand side:

\(\displaystyle \L (\,cot(x)\,+\,csc(x)\,)^{2}\)

\(\displaystyle \L (\,\frac{cos(x)}{sin(x)}\,+\,\frac{1}{sin(x)}\,)^{2}\)

\(\displaystyle \L (\,\frac{cos(x)\,+\,1}{sin(x)}\,)^{2}\)

\(\displaystyle \L \frac{(cos(x)\,+\,1)^{2}}{sin^{2}x}\)


Can you help me to keep going? I am pretty sure this will work.... (in fact) it is identical to the other side with which you did the process to achieve a simplication down to where I am right now....so I know I am /almost/ there.

John.

 

[galactus]

Hey JW:

You are up to:

\(\displaystyle \L\\\frac{(cos(x)+1)^{2}}{1-cos^{2}(x)}\)

=\(\displaystyle \L\\\frac{-2}{cos(x)-1}-1\)

=\(\displaystyle \L\\\frac{-cos(x)-1}{cos(x)-1}\)

=\(\displaystyle \L\\\frac{sec(x)+1}{sec(x)-1}\)

 

[jwpaine]

Hey Galactus,

Hey JW:

You are up to:

\(\displaystyle \L\\\frac{(cos(x)+1)^{2}}{1-cos^{2}(x)}\)

=\(\displaystyle \L\\\frac{-2}{cos(x)-1}-1\) Where did the -2 come from?

=\(\displaystyle \L\\\frac{-cos(x)-1}{cos(x)-1}\)

=\(\displaystyle \L\\\frac{sec(x)+1}{sec(x)-1}\)

 

[galactus]

That's just long division.

Let u=cos(x)