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

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(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

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

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(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

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