The value of tan left(1^{circ}right)+tan left | vedclass

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

Question

(B) Given expression: $	an \left(1^{\circ}ight)+	an \left(89^{\circ}ight)$ Since $	an \left(90^{\circ}-	hetaight)=\cot 	heta$,we have $	an

Vedclass · Accepted Answer

$\frac{2}{\sin \left(2^{\circ}\right)}$

Vedclass · Answer

(B) Given expression: $	an \left(1^{\circ}ight)+	an \left(89^{\circ}ight)$ Since $	an \left(90^{\circ}-	hetaight)=\cot 	heta$,we have $	an \left(89^{\circ}ight)=	an \left(90^{\circ}-1^{\circ}ight)=\cot \left(1^{\circ}ight)$ So,the expression becomes: $	an \left(1^{\circ}ight)+\cot \left(1^{\circ}ight)$ $= \frac{\sin \left(1^{\circ}ight)}{\cos \left(1^{\circ}ight)}+\frac{\cos \left(1^{\circ}ight)}{\sin \left(1^{\circ}ight)}$ $= \frac{\sin^2 \left(1^{\circ}ight)+\cos^2 \left(1^{\circ}ight)}{\sin \left(1^{\circ}ight) \cos \left(1^{\circ}ight)}$ $= \frac{1}{\sin \left(1^{\circ}ight) \cos \left(1^{\circ}ight)}$ Multiply numerator and denominator by $2$: $= \frac{2}{2 \sin \left(1^{\circ}ight) \cos \left(1^{\circ}ight)}$ Using the identity $\sin \left(2	hetaight)=2 \sin 	heta \cos 	heta$: $= \frac{2}{\sin \left(2^{\circ}ight)}$

The value of $\tan \left(1^{\circ}\right)+\tan \left(89^{\circ}\right)$ is

Similar Questions

The expression $[1 - \sin(3\pi - \alpha) + \cos(3\pi + \alpha)] [1 - \sin(\frac{3\pi}{2} - \alpha) + \cos(\frac{5\pi}{2} - \alpha)]$ when simplified reduces to:

$\sin \frac{\pi}{5} + \sin \frac{2\pi}{5} + \sin \frac{3\pi}{5} + \sin \frac{4\pi}{5} =$

The value of $\frac{\sin 55^{\circ} - \cos 55^{\circ}}{\sin 10^{\circ}}$ is

If $\operatorname{cosech} x = \frac{4}{5}$,then $\sinh x = $

Convert $40^{\circ} 20^{\prime}$ into radian measure.

Vedclass Test Series

Exam Paper Generator

Online Exam Module