The value of sqrt{3} operatorname{cosec} 20^{ | vedclass

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

Question

(B) Given expression is $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$.$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} =

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given expression is $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ}$.$= \frac{\sqrt{3}}{\sin 20^{\circ}} - \frac{1}{\cos 20^{\circ}} = \frac{\sqrt{3} \cos 20^{\circ} - \sin 20^{\circ}}{\sin 20^{\circ} \cos 20^{\circ}}$Multiply the numerator and denominator by $2$:$= \frac{2 [(\frac{\sqrt{3}}{2}) \cos 20^{\circ} - (\frac{1}{2}) \sin 20^{\circ}]}{\sin 20^{\circ} \cos 20^{\circ}}$Using the identity $\sin 2	heta = 2 \sin 	heta \cos 	heta$,we have $\sin 20^{\circ} \cos 20^{\circ} = \frac{1}{2} \sin 40^{\circ}$.Also,$\frac{\sqrt{3}}{2} = \sin 60^{\circ}$ and $\frac{1}{2} = \cos 60^{\circ}$.$= \frac{2 [\sin 60^{\circ} \cos 20^{\circ} - \cos 60^{\circ} \sin 20^{\circ}]}{\frac{1}{2} \sin 40^{\circ}}$Using the formula $\sin(A - B) = \sin A \cos B - \cos A \sin B$:$= \frac{2 \sin(60^{\circ} - 20^{\circ})}{\frac{1}{2} \sin 40^{\circ}} = \frac{2 \sin 40^{\circ}}{\frac{1}{2} \sin 40^{\circ}} = 4$.

The value of $\sqrt{3} \operatorname{cosec} 20^{\circ} - \sec 20^{\circ} = $

Similar Questions

Without using trigonometric tables,evaluate: $\sin 48^{\circ} \sec 42^{\circ} + \cos 48^{\circ} \operatorname{cosec} 42^{\circ} = $

If for all real values of $x$,$\frac{4x^2 + 1}{64x^2 - 96x \sin \alpha + 5} < \frac{1}{32}$,then $\alpha$ lies in the interval

If $(1+\tan ^{2} \theta)=\frac{625}{49}$ and $\theta$ is acute,then what is the value of $\sqrt{\sin \theta+\cos \theta}$?

$A$ circular wire of diameter $10\,cm$ is cut and placed along the circumference of a circle of diameter $1\,m$. The angle subtended by the wire at the centre of the circle is equal to

The expression $\frac{\cos 6x + 6\cos 4x + 15\cos 2x + 10}{\cos 5x + 5\cos 3x + 10\cos x}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module