How to prove [?] trigonometric identity with sine and cosine fractions

Key concept: These identities usually look messy until you rewrite everything in sine and cosine. Then the cancellation becomes much more visible.

Step by step

We want to show that

$\frac{1-\cos x}{\sin x}+\frac{\sin x}{1-\cos x}=2\csc x.$

Step 1: Rationalize the first term

Multiply the first fraction by $\frac{1+\cos x}{1+\cos x}$:

$\frac{1-\cos x}{\sin x}$

is already usable, so leave it for now and simplify the second term instead.

Step 2: Use the Pythagorean identity

For the second fraction,

\frac{\sin x(1+\cos x)}{1-\cos^2 x}.$$ Since $$1-\cos^2 x=\sin^2 x,$$ this becomes $$\frac{\sin x(1+\cos x)}{\sin^2 x}=\frac{1+\cos x}{\sin x}.$$ ### Step 3: Add the two parts Now the left-hand side is $$\frac{1-\cos x}{\sin x}+\frac{1+\cos x}{\sin x}= \frac{2}{\sin x}=2\csc x.$$ So the identity is true. ### Final result $$\frac{1-\cos x}{\sin x}+\frac{\sin x}{1-\cos x}=2\csc x.$$ ### Pitfall alert The big trap is forgetting that $1-\cos x=0$ when $x=2k\pi$. At those values, the expression is undefined, so the identity should be interpreted only where both sides are defined. Also, do not cancel terms too early unless you have checked that the denominator is nonzero. ### Try different conditions If the second term were instead $$\frac{\sin x}{1+\cos x},$$ a similar trick works, but you would rationalize with $1-\cos x$ instead. The shape of the proof is the same: rewrite the fraction so the denominator becomes $\sin^2 x$, then combine terms over a common denominator. ### Further reading trigonometric identity, rationalization, Pythagorean identity