Question

# The value of $\tan {5^0}\tan {10^0}\tan {15^0} \cdot \cdot \cdot \cdot \tan {85^0}$ is${\text{A}}{\text{. }}0 \\ {\text{B}}{\text{. Not Defined}} \\ {\text{C}}{\text{. }}1 \\ {\text{D}}{\text{. }} - 1 \\$

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Hint: In this question we need to find the value of the given trigonometric expression. In order to evaluate it easily we will use the property that $\tan x = \cot \left( {90 - x} \right)$. This will simplify the expression and help us reach the answer.

We have been given the expression $\tan {5^0}\tan {10^0}\tan {15^0} \cdot \cdot \cdot \cdot \tan {85^0}$
Now, as we know that $\tan x = \cot \left( {90 - x} \right)$, so we will apply it to the expression starting from $\tan {50^0}$ to $\tan {85^0}$ we get,
$\tan {5^0}\tan {10^0}\tan {15^0} \cdot \cdot \tan {40^0}\tan {45^0}\cot {40^0}...\cot {5^0}$
As we know that, $\tan x.\cot x = 1$
$\Rightarrow \tan {5^0}\tan {10^0}\tan {15^0} \cdot \cdot \cdot \cdot \tan {85^0} = 1 \cdot 1 \cdot 1 \cdot \cdot \cdot \cdot \cdot \tan {45^0}$
And as we know that$\ tan{45^0}=1$,
Hence, $\Rightarrow \tan {5^0}\tan {10^0}\tan {15^0} \cdot \cdot \cdot \cdot \tan {85^0} = 1$