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

Evaluate: $\sin ^2\frac{\pi }{8} + \sin ^2\frac{3\pi }{8} + \sin ^2\frac{5\pi }{8} + \sin ^2\frac{7\pi }{8} = $

The value of $\tan 1^\circ \tan 2^\circ \tan 3^\circ \dots \tan 89^\circ$ is equal to

$\frac{\sin ^2(-160^{\circ})}{\sin ^2 70^{\circ}} + \frac{\sin (180^{\circ}-\theta)}{\sin \theta} = $

$\frac{1-2(\cos 60^{\circ}-\cos 80^{\circ})}{2 \sin 10^{\circ}} = \dots$

$\tan A = \frac{-60}{11}$ and $A$ does not lie in the $4^{\text{th}}$ quadrant. $\sec B = \frac{41}{9}$ and $B$ does not lie in the $1^{\text{st}}$ quadrant. If $\operatorname{cosec} A + \cot B = K$,then $24K =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module