Question

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

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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.

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.