## Trigonometry puzzle

Prove that $1-\cot 23\deg=\frac{2}{1-\cot 22\deg}$

Proof:

To prove the above is the same as proving $(1-\cot 23\deg)(1-\cot 22\deg)=2$

Indeed, by the addition and subtraction formulae, we obtain:

$(1-\cot 23\deg)(1-\cot 22\deg)=(1-\frac{\cos 23\deg}{\sin 23\deg})(1-\frac{\cos 22\deg}{\sin 22\deg})$

which equals $(\frac{\sin 23\deg}-\cos 23\deg{\sin 23\deg})(\frac{\sin 22\deg-\cos 22\deg}{\sin 22\deg})$, which in turn equals

$(\frac{\sqrt{2}\sin(23\deg-45\deg)}{\sin 23\deg-\sin 22\deg})(\frac{\sqrt{2}\sin(22\deg-45\deg)}{1})$, that is,

$\frac{2\sin(-22\deg)\sin (-23\deg)}{\sin(23\deg)\sin (22\deg)}=2$.

Thus, the proof.

More later,

Nalin Pithwa

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