frac{cos 10^{circ} + cos 80^{circ}}{sin 80^{c | vedclass

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(B) Using the sum-to-product formulas: $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$ and $\sin A - \sin B

Vedclass · Accepted Answer

$	an 55^{\circ}$

Vedclass · Answer

(B) Using the sum-to-product formulas: $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} ight) \cos \left( \frac{A-B}{2} ight)$ and $\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} ight) \sin \left( \frac{A-B}{2} ight)$.Applying these to the expression:$\frac{\cos 10^{\circ} + \cos 80^{\circ}}{\sin 80^{\circ} - \sin 10^{\circ}} = \frac{2 \cos \left( \frac{80^{\circ} + 10^{\circ}}{2} ight) \cos \left( \frac{80^{\circ} - 10^{\circ}}{2} ight)}{2 \cos \left( \frac{80^{\circ} + 10^{\circ}}{2} ight) \sin \left( \frac{80^{\circ} - 10^{\circ}}{2} ight)}$$= \frac{\cos 45^{\circ} \cos 35^{\circ}}{\cos 45^{\circ} \sin 35^{\circ}}$$= \cot 35^{\circ} = 	an(90^{\circ} - 35^{\circ}) = 	an 55^{\circ}$.

$\frac{\cos 10^{\circ} + \cos 80^{\circ}}{\sin 80^{\circ} - \sin 10^{\circ}} = ?$

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