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Question

Mathematics Question on Trigonometric Functions

The value of tan⁡(1∘)+ tan(89∘) \tan( 1^\circ) +\, tan(89^\circ) is

A

1sin(1∘)\frac {1} {sin( 1^\circ)}

B

2sin(2∘)\frac {2} {sin( 2^\circ)}

C

2sin(1∘)\frac {2} {sin( 1^\circ)}

D

1sin(2∘)\frac {1} {sin( 2^\circ)}

Answer

2sin(2∘)\frac {2} {sin( 2^\circ)}

Explanation

Solution

tan⁡(1∘)+tan⁡(89∘)=tan⁡(1∘)+tan⁡(90∘−1∘)\tan \left(1^{\circ}\right)+\tan \left(89^{\circ}\right)=\tan \left(1^{\circ}\right)+\tan \left(90^{\circ}-1^{\circ}\right)
=tan⁡(1∘)+cot⁡(1∘)=\tan \left(1^{\circ}\right)+\cot \left(1^{\circ}\right)
=sin⁡(1∘)cos⁡(1∘)+cos⁡(1∘)sin⁡(1∘)=\frac{\sin \left(1^{\circ}\right)}{\cos \left(1^{\circ}\right)}+\frac{\cos \left(1^{\circ}\right)}{\sin \left(1^{\circ}\right)}
=sin⁡2(1∘)+cos⁡2(1∘)cos⁡(1∘)sin⁡(1∘)=\frac{\sin ^{2}\left(1^{\circ}\right)+\cos ^{2}\left(1^{\circ}\right)}{\cos \left(1^{\circ}\right) \sin \left(1^{\circ}\right)}
=1sin⁡(1∘)cos⁡(1∘)=\frac{1}{\sin \left(1^{\circ}\right) \cos \left(1^{\circ}\right)}
=22sin⁡(1∘)cos⁡(1∘)=2sin⁡(2∘)=\frac{2}{2 \sin \left(1^{\circ}\right) \cos \left(1^{\circ}\right)}=\frac{2}{\sin \left(2^{\circ}\right)}