Question

Find the value of $\tan {1^0}\tan {2^0}\tan {3^0}.......\tan {89^0}$.

Answer 1

In the question, we need to determine the value of the function of the series of the tangent which will be not possible but substituting their individual values. So, we need to carry out the process by using some trigonometric identities such as $\tan \theta = \dfrac{1}{{\cot \theta }}$ and $\tan (90 - \theta ) = \cot \theta$.Using the above two formulae, the above question can be solved.

Complete step-by-step solution
Consider the value of the given function to be F such that, $F = \tan {1^0}\tan {2^0}\tan {3^0}........\tan {87^0}\tan {88^0}\tan {89^0}$
As, $\tan (90 - \theta ) = \cot \theta$
Now, the values of the angle of the tangent after ${45^ \circ }$can be written in the form of ${90^ \circ }$ as:

$\Rightarrow \tan 46 = \tan \left( {90 - 44} \right) = \cot 44 \\\ \Rightarrow \tan 47 = \tan \left( {90 - 43} \right) = \cot 43 \\\ \Rightarrow \tan 48 = \tan \left( {90 - 42} \right) = \cot 42 \\\ ... \\\ ... \\\ \Rightarrow \tan 87 = \tan \left( {90 - 3} \right) = \cot 3 \\\ \Rightarrow\tan 88 = \tan \left( {90 - 2} \right) = \cot 2 \\\ \Rightarrow \tan 89 = \tan \left( {90 - 1} \right) = \cot 1 \\\$
Substituting the values in the function $\tan {1^0}\tan {2^0}\tan {3^0}.......\tan {87^0}\tan {88^0}\tan {89^0}$, we get:

$$F = \tan {1^0}\tan {2^0}\tan {3^0}....\tan {45^0}...\tan {87^0}\tan {88^0}\tan {89^0} \\\ = \tan {1^0}\tan {2^0}\tan {3^0}....\tan {45^0}....\tan {(90 - 3)^0}\tan {(90 - 2)^0}\tan {(90 - 1)^0} \\\ = \tan {1^0}\tan {2^0}\tan {3^0}....\tan {45^0}....\cot {3^0}\cot {2^0}\cot {1^0} \\\ = \tan {1^0}\tan {2^0}\tan {3^0}....\tan {45^0}....\dfrac{1}{{\tan {3^0}}}\dfrac{1}{{\tan {2^0}}}\dfrac{1}{{\tan {1^0}}} \\\$$

We can cancel the terms in such a way that every value of $\tan$ up to ${44^0}$ will be canceled by $\cot {44^0}$and only $\tan {45^0}$ is left.

$$F = \tan {1^0}\tan {2^0}\tan {3^0}....\tan {45^0}....\dfrac{1}{{\tan {3^0}}}\dfrac{1}{{\tan {2^0}}}\dfrac{1}{{\tan {1^0}}} \\\ = \tan {45^0} \\\$$

Also, $\tan {45^ \circ } = 1$
Hence, $F = \tan {45^0} = 1$

Hence, $\tan {1^0}\tan {2^0}\tan {3^0}.......\tan {89^0}$ will be equal to 1.

Note: Students should not get confused with the terms degree and radians. These two terms are completely different. To change degrees to radians, the equivalent relationship $1^\circ = \pi /180$radians is used, and the given number of degrees is multiplied by π/180 to convert to radian measure. Similarly, the equation $1^\circ = \pi /180$ is used to change radians to degrees by multiplying the given radian measure $180/\pi$ to obtain the degree measure.