The value of cot 70^circ + 4cos 70^circ is | vedclass

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

Question

(B) Given expression: $\cot 70^\circ + 4\cos 70^\circ$$= \frac{\cos 70^\circ}{\sin 70^\circ} + 4\cos 70^\circ$$= \frac{\cos 70^\circ + 4\sin 70^\circ

Vedclass · Accepted Answer

$\sqrt{3}$

Vedclass · Answer

(B) Given expression: $\cot 70^\circ + 4\cos 70^\circ$$= \frac{\cos 70^\circ}{\sin 70^\circ} + 4\cos 70^\circ$$= \frac{\cos 70^\circ + 4\sin 70^\circ \cos 70^\circ}{\sin 70^\circ}$$= \frac{\cos 70^\circ + 2\sin 140^\circ}{\sin 70^\circ}$$= \frac{\cos 70^\circ + 2\sin(180^\circ - 40^\circ)}{\sin 70^\circ}$$= \frac{\cos 70^\circ + 2\sin 40^\circ}{\sin 70^\circ}$$= \frac{\sin 20^\circ + \sin 40^\circ + \sin 40^\circ}{\sin 70^\circ}$$= \frac{\sin 20^\circ + 2\sin 40^\circ}{\cos 20^\circ}$$= \frac{\sin 20^\circ + 2(2\sin 20^\circ \cos 20^\circ)}{\cos 20^\circ}$ (This approach is complex,let's use the standard identity approach:)$= \frac{\cos 70^\circ + 2\sin 140^\circ}{\sin 70^\circ} = \frac{\sin 20^\circ + 2\sin 40^\circ}{\cos 20^\circ} = \frac{\sin 20^\circ + \sin 40^\circ + \sin 40^\circ}{\cos 20^\circ} = \frac{2\sin 30^\circ \cos 10^\circ + \sin 40^\circ}{\cos 20^\circ}$$= \frac{\cos 10^\circ + \sin 40^\circ}{\cos 20^\circ} = \frac{\sin 80^\circ + \sin 40^\circ}{\cos 20^\circ} = \frac{2\sin 60^\circ \cos 20^\circ}{\cos 20^\circ} = 2 	imes \frac{\sqrt{3}}{2} = \sqrt{3}$.

The value of $\cot 70^\circ + 4\cos 70^\circ$ is

Similar Questions

If $A$ lies in the third quadrant and $3 \tan A - 4 = 0$,then find the value of $5 \sin 2A + 3 \sin A + 4 \cos A$.

Which one of the following is true?

If $A, B, C$ are angles of a triangle,then $\sin 2A + \sin 2B - \sin 2C$ is equal to

The expression $\frac{\tan 57^{\circ}+\cot 37^{\circ}}{\tan 33^{\circ}+\cot 53^{\circ}}$ is equal to

If $\cos x + \cos^2 x = 1$,then the value of $\sin^2 x + \sin^4 x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module