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) સરવાળા-ગુણાકારના સૂત્રોનો ઉપયોગ કરતા: $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} \right) \cos \left( \frac{A-B}{2} \right)$ અને $\sin A - \sin

Vedclass · Accepted Answer

$	an 55^{\circ}$

Vedclass · Answer

(B) સરવાળા-ગુણાકારના સૂત્રોનો ઉપયોગ કરતા: $\cos A + \cos B = 2 \cos \left( \frac{A+B}{2} ight) \cos \left( \frac{A-B}{2} ight)$ અને $\sin A - \sin B = 2 \cos \left( \frac{A+B}{2} ight) \sin \left( \frac{A-B}{2} ight)$.આ સૂત્રોનો ઉપયોગ કરતા:$\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

સાબિત કરો કે $2 \sin ^{2} \frac{3 \pi}{4} + 2 \cos ^{2} \frac{\pi}{4} + 2 \sec ^{2} \frac{\pi}{3} = 10$.

$\tan \theta \sin \left( \frac{\pi }{2} + \theta \right) \cos \left( \frac{\pi }{2} - \theta \right) = $

$\cos 1^{\circ} \cdot \cos 2^{\circ} \cdot \cos 3^{\circ} \dots \cos 179^{\circ} = $

$\cos 255^o + \sin 195^o$ ની કિંમત શોધો.

જો $x \sin 45^\circ \cos^2 60^\circ = \frac{\tan^2 60^\circ \csc 30^\circ}{\sec 45^\circ \cot^2 30^\circ}$ હોય,તો $x = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module