prove that ((1−cosA+cosB−cos(A+B))/(1+cosA−cosB−cos(A+B)))=tan(A/2)∙cot(B/2)


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Question Number 201425 by mathlove last updated on 06/Dec/23

$p r o v e t h a t$ $\frac{1 - c o s A + c o s B - c o s (A + B)}{1 + c o s A - c o s B - c o s (A + B)} = t a n \frac{A}{2} \cdot c o t \frac{B}{2}$
	Answered by esmaeil last updated on 06/Dec/23

	$\frac{2 {s i n}^{2} (\frac{A + B}{2}) - 2 s i n (\frac{B + A}{2}) s i n (\frac{B - A}{2})}{2 {s i n}^{2} (\frac{B + A}{2}) - 2 s i n (\frac{B + A}{2}) s i n (\frac{A - B}{2})} =$ $\frac{s i n \frac{B + A}{2} + s i n (\frac{A - B}{2})}{s i n (\frac{B + A}{2}) - s i n (\frac{A - B}{2})} = \frac{2 s i n \frac{A}{2} c o s \frac{B}{2}}{2 s i n \frac{B}{2} c o s \frac{A}{2}}$ $t a n \frac{A}{2} c o t \frac{B}{2} (1 - c o s (A B + A), c o s B - c o s A) .$
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