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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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}$.

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

Similar Questions

If $\sec \theta = \frac{13}{12}$ and $\theta$ lies in the $4^{\text{th}}$ quadrant,then $\tan \theta \times \operatorname{cosec} \theta \times \sin \theta \times \cos \theta = $

Find the value of the trigonometric function $\sin 765^{\circ}$.

$\cos A + \sin (270^\circ + A) - \sin (270^\circ - A) + \cos (180^\circ + A) = $

$\cos ^2\left(\frac{7 \pi}{8}\right)+\cos ^2\left(\frac{5 \pi}{8}\right)+\cos ^2\left(\frac{3 \pi}{8}\right)+\cos ^2\left(\frac{\pi}{8}\right)=$

Find the value of the trigonometric function $\cot \left(-\frac{15 \pi}{4}\right)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module