Chapter 11: Problem 89
$$ \frac{\sin A-\sin B}{\cos B-\cos A}=\cot \frac{A+B}{2} $$
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Chapter 11: Problem 89
$$ \frac{\sin A-\sin B}{\cos B-\cos A}=\cot \frac{A+B}{2} $$
These are the key concepts you need to understand to accurately answer the question.
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$$ 2 \cos \frac{\pi}{13} \cos \frac{9 \pi}{13}+\cos \frac{3 \pi}{13}+\cos \frac{5 \pi}{13}=0 $$
$$ \text { If } x=y \cos \frac{2 \pi}{3}=z \cos \frac{4 \pi}{3}, \text { then show that } x y+y z+z x=0 $$
$$ (1+\cot A+\tan A)(\sin A-\cos A)=\frac{\sec A}{\operatorname{cosec}^{2} A}-\frac{\cos e c A}{\sec ^{2} A} $$
$$ \sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ}=\frac{1}{8} $$
$$ \sqrt{\operatorname{cosec}^{2} A-1}=\cos A \operatorname{cosec} A $$
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