The exercise is to prove the trig identity by rewriting each side of the equation into the same form. However only the following identities can be used in the process: $$\begin{align*} \tan \theta &= \frac{\sin \theta}{\cos \theta}\\ (\sin \theta)^2 + (\cos \theta)^2 &= 1\\ \csc \theta &= \frac{1}{\sin \theta}\\ \sec \theta &= \frac{1}{\cos \theta}\\ \cot \theta &= \frac{1}{\tan \theta}. \end{align*}$$

The trig identity given is:

$$\sin x + \sin x \cdot \cot^2 x = \csc x $$

I simplified it to:

$$\sin x + \sin x \cdot \left(\frac{\cos x}{\sin x}\right)^2 = \frac{1}{\sin x} $$

After which I get lost in a mess of term rewriting that never seems to lead anywhere fruitful. Any hints would be greatly appreciated.

  • 3
    $\begingroup$ Multiply both sides by $\sin x$ and see what you get. $\endgroup$ Jun 1, 2012 at 19:03
  • $\begingroup$ Arturo, you are simply too quick at editing. You finished mere seconds before I was done. $\endgroup$
    – Joe
    Jun 1, 2012 at 19:04
  • $\begingroup$ you are at last step: $\frac {1}{sinx}$=Cosecx $\endgroup$ May 10, 2013 at 4:41

2 Answers 2


Start with $\sin x + \sin x\cot^2 x$. Factoring out $\sin x$ we get $$\sin x + \sin x\cot^2 x = \sin x (1 + \cot^2 x).$$ Now use the definition of $\cot x$ and rewrite the expression in the parenthesis as a single fraction: $$\begin{align*} \sin x + \sin x\cot^2x &= \sin x\Bigl( 1 + \cot^2 x\Bigr)\\ &= \sin x \left( 1 + \frac{\cos^2 x}{\sin^2 x}\right)\\ &= \sin x \left(\frac{\sin^2 x}{\sin ^2x} + \frac{\cos^2 x}{\sin^2x}\right)\\ &= \sin x\left(\frac{\sin^2 x + \cos^2x}{\sin^2x}\right)\\ &= \sin x\left(\frac{1}{\sin^2 x}\right). \end{align*}$$

Note that I am not manipulating the identity as a whole. One has to be careful when doing that, because you can start with a false equality and end with a true one if not all your steps are reversible. Instead, we start with one side, and apply known identities to it until it becomes the other side.

(I am guessing you can take it home from where I left it, of course...)

  • $\begingroup$ Very elaborate and rigorous. Great answer. $+1$ $\endgroup$
    – Mr Pie
    Mar 2, 2021 at 23:25

Taking the left hand side, try cancelling the $\sin^2{x}$ with the $\sin{x}$ in the numerator

(i.e. $\sin{x} .(\frac{\cos^2{x}}{\sin^2{x}}) = \frac{\cos^2{x}}{\sin{x}}$)

Then add the remaining two terms ($\sin{x} + \frac{\cos^2{x}}{\sin{x}}$) and use the identity $\sin^2{x}+\cos^2{x}=1$. That should do the trick :)


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