The problem is:
Prove the identity $$ \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=\csc(x)-\cot(x)$$
The text's solution goes as follows:
$$ \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}= \sqrt{\frac{(1-\cos(x))^2}{1-\cos^2(x)}}= \frac{1-\cos(x)}{\sin(x)}= \csc(x)-\cot(x).
$$
My own solution is almost the same. However, in the third equality above, I include the other solution:
$$
\frac{1-\cos(x)}{-\sin(x)} = \cot(x)-\csc(x).
$$
I made the graphs of both sides of the identity and found out that each of the two solutions on the R.H.S. actually only matches with the L.H.S. for half of the domain. Thus, I conclude that:
- The problem's statement is just totally wrong.
- To make the problem sensible, I should specify the domain on which the R.H.S is well-defined(i.e. excluding the points where $\sin(x)=0$, and is positive. So, for the solution $$\csc(x)-\cot(x)\enspace ,$$ the domain should be $ \cup_{k \in \mathbb{Z}} ( \,2k\pi,\,(2k+1)\pi)$.
Since I have to explain the problem and the solution to my younger sister, I'd like to be sure my conclusions and understanding are right. Please give me some advice. Thanks.
Best Answer
We must always remember that $\sqrt {x^2}=|x|$
However if we know that $x<0$, then $\sqrt {x^2}=-x$ and if $x>0$, then $\sqrt {x^2}=x$
In your question , as illustrated in comments the correct answer should be
$$ \sqrt{\frac{1-\cos(x)}{1+\cos(x)}}=|\csc(x)-\cot(x).|$$
Now if we are given the domain of $x$, then we can further expand it
$For Ex.$ If $x \in (\frac{3\pi}{2},2\pi)$, then RHS would be $-(\csc(x)-\cot(x))$