sqrt{3} csc 20^circ - sec 20^circ = | vedclass

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

Question

(C) $\sqrt{3} \csc 20^\circ - \sec 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}$ $= \frac{\sqrt{3} \cos 20^\circ - \sin 20^\cir

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) $\sqrt{3} \csc 20^\circ - \sec 20^\circ = \frac{\sqrt{3}}{\sin 20^\circ} - \frac{1}{\cos 20^\circ}$ $= \frac{\sqrt{3} \cos 20^\circ - \sin 20^\circ}{\sin 20^\circ \cos 20^\circ}$ અંશ અને છેદને $2$ વડે ગુણતા: $= \frac{2 \left( \frac{\sqrt{3}}{2} \cos 20^\circ - \frac{1}{2} \sin 20^\circ \right)}{\frac{1}{2} (2 \sin 20^\circ \cos 20^\circ)}$ $= \frac{2 (\sin 60^\circ \cos 20^\circ - \cos 60^\circ \sin 20^\circ)}{\frac{1}{2} \sin 40^\circ}$ $= \frac{2 \sin(60^\circ - 20^\circ)}{\frac{1}{2} \sin 40^\circ} = \frac{2 \sin 40^\circ}{\frac{1}{2} \sin 40^\circ} = 4$.

$\sqrt{3} \csc 20^\circ - \sec 20^\circ = $

Similar Questions

જો $\sin \theta_1 + \sin \theta_2 + \sin \theta_3 = 3$ હોય,તો $\cos \theta_1 + \cos \theta_2 + \cos \theta_3 = $

જો $\tan A + \cot A = 2$ હોય,તો $\tan^{10} A + \cot^{10} A$ ની કિંમત શોધો.

$\left( \frac{\sin 2A}{1 + \cos 2A} \right) \left( \frac{\cos A}{1 + \cos A} \right) = $

$E = \frac{25 \sec^4 x - 50 \sec^2 x + 74}{\tan^2 x}$ ની ન્યૂનતમ કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module