$$\frac {(\cot \theta + 1)(\cot \theta + 1)-\csc^2 \theta}{\cot \theta}$$

Here's what I have

$$\csc^2 \theta - \cot^2 \theta =1 \qquad\text{(Pythagorean Identity)}$$

Then, I consolidate the numerator and rewrite the equation:

$$\frac {\cot^2 \theta + 1-\csc^2 \theta}{\cot \theta}$$

The next step is where I believe I've gone wrong:

$$\cot \theta + 1-\csc^2 \theta$$

Rewritten as:

$$\csc^2 \theta - \cot \theta = 1$$

I'm not confident that this is correct, but if it is, do I need to simplify further?

Note: I tried solving this by replacing $\cot$ and $\csc$ with $\frac{\cos}{\sin}$ and $\frac{1}{\sin}$, respectively, but I hit a wall midway through. Any advice is valued.

Edit: The original post had a transcription error. The expression should be reducible now.

  • 1
    $\begingroup$ Is the original expression written correctly? $\endgroup$ – mrnovice Mar 28 '17 at 22:29
  • $\begingroup$ How do you get $\cot^2\theta+1$ from $(\cos\theta+1)^2$? $\endgroup$ – egreg Mar 28 '17 at 22:32
  • $\begingroup$ @egreg I guess my logic was that 1^2 is still 1, but I suppose how I wrote it is still incorrect due to FOIL. $\endgroup$ – JCD Mar 28 '17 at 22:34
  • 1
    $\begingroup$ @JCD $(\cos\theta+1)^2=\cos^2\theta+2\cos\theta+1$. But is it $\cos\theta+1$ or $\cot\theta+1$? $\endgroup$ – egreg Mar 28 '17 at 22:41
  • 1
    $\begingroup$ Correct your expansion, and apply the trig identity to $\csc^2(x)$, it should work out. $\endgroup$ – Kaynex Mar 28 '17 at 22:57

Your expansion seems to be wrong. After expanding, you should have: $$\frac{(\cot{\theta}+1)^2-\csc^2{\theta}}{\cot{\theta}}=\frac{\cot^2{\theta}+2\cot{\theta}+\color{blue}{1-\csc^2{\theta}}}{\cot{\theta}}$$ Now, use the identity that you've mentioned on the $\color{blue}{\text{blue}}$ text: $$\csc^2{\theta}-\cot^2{\theta}\equiv 1$$ And then everything should cancel out very nicely.

  • $\begingroup$ Continuing from where you left off, am I on the right track? $\ 2 + cot \theta + \dfrac{\ 1 - csc^2 \theta}{\cot \theta}$ ..... $\ 2cot\theta +cot^2 \theta + 1-csc^2 $ ..... $\ 1-csc^2\theta +cot^2 \theta + 2cot \theta $ ...... $\ -csc^2\theta +cot^2 \theta + 2cot \theta = -1$ ...... $\ csc^2\theta -cot^2 \theta - 2cot \theta = 1$ If i'm correct thus far, then cot must be zero, which makes the entire equation ..undefined? $\endgroup$ – JCD Mar 29 '17 at 0:50
  • $\begingroup$ I'm sorry for the formatting. I'm new to this and apparently comments don't support /newlines $\endgroup$ – JCD Mar 29 '17 at 0:52
  • $\begingroup$ @projectilemotion, Why don't u apply the identity in the second line? $\endgroup$ – lab bhattacharjee Mar 29 '17 at 1:15
  • $\begingroup$ @labbhattacharjee Oh yes, you are right. Thank you for the feedback! $\endgroup$ – projectilemotion Mar 29 '17 at 5:52
  • $\begingroup$ @JCD No, that is not correct. You should get $2$ as an answer. If you rearrange the identity, you have: $$1-\csc^2{\theta} \equiv -\cot^2{\theta}$$ Now apply this identity on the blue text. $\endgroup$ – projectilemotion Mar 29 '17 at 5:58

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