sqrt{3} cot 20^{circ} - 4 cos 20^{circ} નું મ | vedclass

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

Question

(A) ધારો કે $E = \sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ}$.$E = \sqrt{3} \frac{\cos 20^{\circ}}{\sin 20^{\circ}} - 4 \cos 20^{\circ}$$E = \frac{\s

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) ધારો કે $E = \sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ}$.$E = \sqrt{3} \frac{\cos 20^{\circ}}{\sin 20^{\circ}} - 4 \cos 20^{\circ}$$E = \frac{\sqrt{3} \cos 20^{\circ} - 4 \sin 20^{\circ} \cos 20^{\circ}}{\sin 20^{\circ}}$$2 \sin 	heta \cos 	heta = \sin 2	heta$ નો ઉપયોગ કરતા,$4 \sin 20^{\circ} \cos 20^{\circ} = 2 \sin 40^{\circ}$ મળે.$E = \frac{\sqrt{3} \cos 20^{\circ} - 2 \sin 40^{\circ}}{\sin 20^{\circ}}$$E = \frac{\sqrt{3} \cos 20^{\circ} - 2 \sin(60^{\circ} - 20^{\circ})}{\sin 20^{\circ}}$$E = \frac{\sqrt{3} \cos 20^{\circ} - 2 (\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ})}{\sin 20^{\circ}}$$E = \frac{\sqrt{3} \cos 20^{\circ} - 2 (\frac{\sqrt{3}}{2} \cos 20^{\circ} - \frac{1}{2} \sin 20^{\circ})}{\sin 20^{\circ}}$$E = \frac{\sqrt{3} \cos 20^{\circ} - \sqrt{3} \cos 20^{\circ} + \sin 20^{\circ}}{\sin 20^{\circ}}$$E = \frac{\sin 20^{\circ}}{\sin 20^{\circ}} = 1$.

$\sqrt{3} \cot 20^{\circ} - 4 \cos 20^{\circ}$ નું મૂલ્ય કેટલું થાય?

Similar Questions

$4 \cos \frac{\pi}{7} \cos \frac{\pi}{5} \cos \frac{2 \pi}{7} \cos \frac{2 \pi}{5} \cos \frac{4 \pi}{7} = $

જો $\cos x + \cos y = -\cos \alpha$ અને $\sin x + \sin y = -\sin \alpha$ હોય,તો $\cot \left(\frac{x+y}{2}\right) = $

$\theta$ નું સૌથી નાનું ધન મૂલ્ય (અંશમાં) જેના માટે $\tan(\theta+100^{\circ})=\tan(\theta+50^{\circ}) \tan(\theta) \tan(\theta-50^{\circ})$ માન્ય છે,તે છે ($^{\circ}$ માં)

જો $|\sin \alpha - \cos \alpha| = \frac{3}{4}$ હોય,તો $|\sec 2\alpha - \tan 2\alpha| = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module