Question

The value of $e^{{\log }_{10}\tan 1^{\circ }+{\log }_{10}\tan 2^{\circ }+{\log }_{10}\tan 3^{\circ }+.......+{\log }_{10}\tan {89}^{\circ }}$is

• A. $0$
• B. $e$
• C. $\frac{1}{3}$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option:

Find the required value.

Given: $e^{{\log }_{10}\tan 1^{\circ }+{\log }_{10}\tan 2^{\circ }+{\log }_{10}\tan 3^{\circ }+.......+{\log }_{10}\tan {89}^{\circ }}$

Let,

$E=e^{{\log }_{10}\tan 1^{\circ }+{\log }_{10}\tan 2^{\circ }+{\log }_{10}\tan 3^{\circ }+.......+{\log }_{10}\tan {89}^{\circ }}$

$$\begin{gathered} \begin{array}{rl}& =e^{{\log }_{10}\tan 1^{\circ }+{\log }_{10}\tan 2^{\circ }+{\log }_{10}\tan 3^{\circ }+.....+{\log }_{10}\tan {45}^{\circ }+{\log }_{10}\cot ({90}^{\circ }-{46}^{\circ })+{\log }_{10}\cot ({90}^{\circ }-{47}^{\circ })+......+{\log }_{10}\cot ({90}^{\circ }-{89}^{\circ })}\left[\because \tan \theta =\cot ({90}^{\circ }-\theta )\right]\\ & =e^{{\log }_{10}\tan 1^{\circ }+{\log }_{10}\tan 2^{\circ }+{\log }_{10}\tan 3^{\circ }+.....+{\log }_{10}\tan {45}^{\circ }+{\log }_{10}\cot {44}^{\circ }+{\log }_{10}\cot {43}^{\circ }+......+{\log }_{10}\cot 1^{\circ }}\\ & =e^{({\log }_{10}\tan 1^{\circ }+{\log }_{10}\cot 1^{\circ })+({\log }_{10}\tan 2^{\circ }+{\log }_{10}\cot 2^{\circ })+({\log }_{10}\tan 3^{\circ }+{\log }_{10}\cot 3^{\circ })+........+({\log }_{10}\tan {44}^{\circ }+{\log }_{10}\cot {44}^{\circ })+{\log }_{10}\tan {45}^{\circ }}\end{array} \\ =e^{({\log }_{10}\tan 1^{\circ }\times \cot 1^{\circ })+({\log }_{10}\tan 2^{\circ }\times \cot 2^{\circ })+({\log }_{10}\tan 3^{\circ }\times \cot 3^{\circ })+........+({\log }_{10}\tan {44}^{\circ }\times \cot {44}^{\circ })+{\log }_{10}\tan {45}^{\circ }}\left[\because {\log }_{a}m+{\log }_{a}n={\log }_{a}m\times n\right] \\ =e^{{\log }_{10}1+{\log }_{10}1+{\log }_{10}1+........+{\log }_{10}1}\left[\because \tan \theta cot\theta =1,\tan {45}^{\circ }=1\right] \\ \begin{array}{rl}& =e^{0+0+0+.....+0}\left[\because {\log }_{10}1=0\right]\\ & =e^0\\ & =1\end{array} \end{gathered}$$
Hence, option (D) is the correct answer.