Question

The value of $\tan (1{}^\circ )+\tan (89{}^\circ )$ is equal to
A. $\dfrac{1}{\sin 1{}^\circ }$
B. $\dfrac{2}{\sin 2{}^\circ }$
C. $\dfrac{2}{\sin 1{}^\circ }$
D. $\dfrac{1}{\sin 2{}^\circ }$
E. $\dfrac{\sin 1{}^\circ }{2}$

Answer 1

Hint: Use the identity $\tan (90-x)=\cot x$ to simplify one of the two, $\tan 89{}^\circ $ and $\tan 1{}^\circ $. Then proceed by simplifying the tan or cot, in terms of sine and cosine of the angles, to arrive at the answer.

We need to find the value of the expression $\tan (1{}^\circ )+\tan (89{}^\circ )$.
Since, in the answers we can’t see any expression that contains $89{}^\circ $, or any multiple of it, we have to simplify $\tan 89{}^\circ $ to arrive at the answer.
To do this, we’ll make use of the formula $\tan (90-x){}^\circ =\cot x{}^\circ $.
We can rewrite $89{}^\circ =(90-1){}^\circ $. Thus, in this case, $x=1$. Applying the formula, we’ll get :
$\begin{align}
  & \tan (90-1){}^\circ =\cot 1{}^\circ \\
 & \Rightarrow \tan 89{}^\circ =\cot 1{}^\circ \\
\end{align}$
Thus, we have simplified the expression a bit. Therefore, the original expression $\tan (1{}^\circ )+\tan (89{}^\circ )$ now becomes equal to : $\begin{align}
  & \tan (1{}^\circ )+\tan (89{}^\circ ) \\
 & =\tan 1{}^\circ +\cot 1{}^\circ \\
\end{align}$
Now, we can see that all the answers are in terms of the sine function.
Thus, it will be wise of us to simplify tan in terms of sin and cos.
We know that, $\begin{align}
  & \tan x{}^\circ =\dfrac{\sin x{}^\circ }{\cos x{}^\circ } \\
 & \cot x{}^\circ =\dfrac{\cos x{}^\circ }{\sin x{}^\circ } \\
\end{align}$
Therefore, we can apply these formulas to simplify the expression we have in terms of sine and cosine.
Therefore, $\begin{align}
  & \tan (1{}^\circ )+\tan (89{}^\circ ) \\
 & =\tan 1{}^\circ +\cot 1{}^\circ \\
 & =\dfrac{\sin 1{}^\circ }{\cos 1{}^\circ }+\dfrac{\cos 1{}^\circ }{\sin 1{}^\circ } \\
 & =\dfrac{{{\sin }^{2}}1{}^\circ +{{\cos }^{2}}1{}^\circ }{\sin 1{}^\circ \cos 1{}^\circ } \\
\end{align}$
Now, we can make use of the identity : ${{\sin }^{2}}x+{{\cos }^{2}}x=1$, and substitute for the numerator that we got on simplifying the expression in terms of sin and cos.
Substituting for ${{\sin }^{2}}1{}^\circ +{{\cos }^{2}}1{}^\circ $, we get :
$\begin{align}
  & \dfrac{{{\sin }^{2}}1{}^\circ +{{\cos }^{2}}1{}^\circ }{\sin 1{}^\circ \cos 1{}^\circ } \\
 & =\dfrac{1}{\sin 1{}^\circ \cos 1{}^\circ } \\
\end{align}$
Now, let’s multiply the numerator and the denominator by 2.
Doing so, we get :
$\dfrac{1}{\sin 1{}^\circ \cos 1{}^\circ }=\dfrac{2}{2\sin 1{}^\circ \cos 1{}^\circ }$
Now, we can identify the RHS of an identity in the denominator.
To refresh your memory, here is the identity : $\sin 2x=2\sin x\cos x$.
Here,$x=1$. Therefore, we know that $2\sin 1{}^\circ \cos 1{}^\circ =\sin (2\times 1){}^\circ =\sin 2{}^\circ $.
Hence, we can now substitute for the denominator, in the expression at arrive at a more simplified expression. Substituting for the denominator, using the identity, we get :
$\dfrac{2}{2\sin 1{}^\circ \cos 1{}^\circ }=\dfrac{2}{\sin 2{}^\circ }$
Hence, we now have the final answer only in terms of sine of some angle. Now, we can compare this answer with the options given to us. Doing so, we find that our required answer is option B. $\dfrac{2}{\sin 2{}^\circ }$.
Hence, our answer is $\dfrac{2}{\sin 2{}^\circ }$.

Note: You could also simplify the term $\tan 1{}^\circ $, and simplify from there on, in terms of sine or cosine of $89{}^\circ $, and once you’re done reducing it to the simplest form of sine and cosine, you could easily replace $89{}^\circ $ in terms of $1{}^\circ $ again. However, this process is more lengthy, and hence, it’s simpler to simplify $\tan 89{}^\circ $ from the start, because we can see the answers and make a guess it’s more beneficial to do so.