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The value of \[\tan ({1^ \circ }) + \tan ({89^ \circ })\]is___
\[
  A.\dfrac{1}{{\sin ({1^ \circ })}} \\
  B.\dfrac{2}{{\sin ({2^ \circ })}} \\
  C.\dfrac{2}{{\sin ({1^ \circ })}} \\
  D.\dfrac{1}{{\sin ({2^ \circ })}} \\
\]

Answer
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Hint: We will use the trigonometric ratios of complementary angles. The complementary angle of \[\tan \theta \] is \[\cot ({90^ \circ } - \theta )\]. We also have to use the trigonometric identity \[\sin 2\theta = 2\sin \theta \cos \theta \] in the further parts of the question.

Complete step-by-step answer:
We are given two trigonometric ratios in the question \[tan({{\text{1}}^ \circ }{\text{) }}and{\text{ }}tan({89^ \circ })\].
We will proceed further by converting \[tan({{\text{1}}^ \circ }{\text{)}}\] into its complementary angle.
\[tan({{\text{1}}^ \circ }{\text{) = cot(90 - 1}}{{\text{)}}^ \circ } = \cot ({89^ \circ })\]
Therefore,
\[
  \tan ({1^ \circ }) + \tan ({89^ \circ }) \\
   = \tan ({90^ \circ } - {89^ \circ }) + \tan ({89^ \circ }) \\
   = \cot ({89^ \circ }) + \tan ({89^ \circ }) \\
\]
Now, we will split the ratios into \[\sin \] and \[\cos \]
\[ \Rightarrow \dfrac{{\cos ({{89}^ \circ })}}{{\sin ({{89}^ \circ })}} + \dfrac{{\sin ({{89}^ \circ })}}{{\cos ({{89}^ \circ })}}\]
We will take the LCM of the two denominators,
\[ \Rightarrow \dfrac{{{{\cos }^2}({{89}^ \circ }) + {{\sin }^2}({{89}^ \circ })}}{{\cos ({{89}^ \circ })\sin ({{89}^ \circ })}}\]
We know that \[{\cos ^2}\theta + {\sin ^2}\theta = 1\], so
\[ \Rightarrow \dfrac{1}{{\cos ({{89}^ \circ })\sin ({{89}^ \circ })}}\]
Now, we will multiply \[2\] in both the numerator and denominator,
\[
   \Rightarrow \dfrac{{2 \times 1}}{{2 \times \cos ({{89}^ \circ })\sin ({{89}^ \circ })}} \\
   = \dfrac{2}{{2\sin ({{89}^ \circ })\cos ({{89}^ \circ })}} \\
\]
We know that \[\sin 2\theta = 2\sin \theta \cos \theta \], so \[2\sin ({89^ \circ })\cos ({89^ \circ }) = \sin 2 \times {89^ \circ }\].
Therefore,
\[
  \dfrac{2}{{2\sin ({{89}^ \circ })\cos ({{89}^ \circ })}} \\
   = \dfrac{2}{{\sin (2 \times {{89}^ \circ })}} \\
   = \dfrac{2}{{\sin ({{178}^ \circ })}} \\
   = \dfrac{2}{{\sin ({{180}^ \circ } - {2^ \circ })}} \\
\]
Angle \[\theta \] lies in the first quadrant, where, \[90^\circ > \theta > 0^\circ \]and \[\left( {180^\circ - \theta } \right)\] lies in the 2nd quadrant. In the first and the second quadrant, \[sin\theta \] is always positive.
So,\[sin\left( {{{180}^ \circ } - \theta } \right) = sin\theta \]
Therefore,
\[
  \dfrac{2}{{\sin ({{180}^ \circ } - {2^ \circ })}} \\
   = \dfrac{2}{{\sin ({2^ \circ })}} \\
\]
\[\therefore \tan (1^\circ ) + \tan (89^\circ ) = \dfrac{2}{{\sin (2^\circ )}}\]
Thus, the answer is option B.

Note: In these types of questions, we need to remember all the trigonometric identities that we have studied. All the trigonometric formulas are very important to solve problems like these.