Question

The value of ${{e}^{{{\log }_{10}}\tan 1+{{\log }_{10}}\tan 2+{{\log }_{10}}\tan 3+.......+{{\log }_{10}}\tan 89}}$ is
(a) 0
(b) 1
(c) $\dfrac{1}{e}$
(d) $e$

Answer 1

Hint: Convert the angles of tangent beyond ${{45}^{\circ }}$ into angles of cotangent using the complementary angle transformation hence, pair first logarithmic term with the last, second logarithmic term with second last and so on. Use the product rule of logarithm to convert into a single term.

Complete step-by-step answer:
Let us come to the question. Let the value of the given expression be ‘$E$’. Therefore,
\[E={{e}^{{{\log }_{10}}\tan 1+{{\log }_{10}}\tan 2+{{\log }_{10}}\tan 3+.......+{{\log }_{10}}\tan 89}}\]
Now, changing the angles of tangent after ${{45}^{\circ }}$ into angles of cotangent using complementary angle rule: $\tan \theta =\cot \left( {{90}^{\circ }}-\theta \right)$, we get,
$\begin{align}
  & E={{e}^{{{\log }_{10}}\tan 1+{{\log }_{10}}\tan 2+{{\log }_{10}}\tan 3+.....+{{\log }_{10}}\tan 45+{{\log }_{10}}\cot (90-46)+{{\log }_{10}}\cot (90-47)+......+{{\log }_{10}}\cot (90-89)}} \\
 & \text{ }={{e}^{{{\log }_{10}}\tan 1+{{\log }_{10}}\tan 2+{{\log }_{10}}\tan 3+.....+{{\log }_{10}}\tan 45+{{\log }_{10}}\cot 44+{{\log }_{10}}\cot 43+......+{{\log }_{10}}\cot 1}} \\
\end{align}$
Now, pairing the logarithmic terms having same angle of tan and cot, we have,
$E={{e}^{({{\log }_{10}}\tan 1+{{\log }_{10}}\cot 1)+({{\log }_{10}}\tan 2+{{\log }_{10}}\cot 2)+({{\log }_{10}}\tan 3+{{\log }_{10}}\cot 3)+........+({{\log }_{10}}\tan 44+{{\log }_{10}}\cot 44)+{{\log }_{10}}\tan 45}}$
Here, we will apply the product rule of logs. It says that logarithm of a product is equal to a sum of logarithms. Mathematically, ${{\log }_{a}}(m+n)={{\log }_{a}}mn$. Therefore,
$E={{e}^{({{\log }_{10}}\tan 1\times \cot 1)+({{\log }_{10}}\tan 2\times \cot 2)+({{\log }_{10}}\tan 3\times \cot 3)+........+({{\log }_{10}}\tan 44\times \cot 44)+{{\log }_{10}}\tan 45}}$
There is a trigonometric identity that: product of tangent of an angle and co-tangent of the same angle results in 1. Mathematically, $\tan \theta \times \cot \theta =1$. Also, we know that, $\tan {{45}^{\circ }}=1$. Using these relations, we have,
$E={{e}^{{{\log }_{10}}1+{{\log }_{10}}1+{{\log }_{10}}1+........+{{\log }_{10}}1}}$
We know that, ${{\log }_{a}}1=0$.
 $\begin{align}
  & \therefore E={{e}^{0+0+0+.....+0}} \\
 & \text{ }={{e}^{0}} \\
 & \text{ }=1 \\
\end{align}$
Hence, option (b) is the correct answer.

Note: Here, all the angles are paired except ${{45}^{\circ }}$ because there are 89 terms of tangent of angle and 44 are paired, leaving 45th angle which is ${{45}^{\circ }}$. It is important to convert the tangents into the cotangents because only then we can apply the product rule in logarithms. Note that the value of $\log 1=0$, provided base of the log is defined.