Chapter 11: Problem 76
$$ \frac{\cos 2 B-\cos 2 A}{\sin 2 B+\sin 2 A}=\tan (A-B) $$
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Chapter 11: Problem 76
$$ \frac{\cos 2 B-\cos 2 A}{\sin 2 B+\sin 2 A}=\tan (A-B) $$
These are the key concepts you need to understand to accurately answer the question.
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$$ \sin 20^{\circ} \sin 40^{\circ} \sin 60^{\circ} \sin 80^{\circ}=\frac{3}{16} $$
$$ \cos 36^{\circ} \cos 42^{\circ} \cos 78^{\circ}=\frac{1}{8} $$
$$ \frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}=2 \operatorname{cosec} A $$
$$ \frac{\sin (4 A-2 B)+\sin (4 B-2 A)}{\cos (4 A-2 B)+\cos (4 B-2 A)}=\tan (A+B) $$
$$ \frac{\operatorname{cosec} A}{\cot A+\tan A}=\cos A $$
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