8. Introduction to Trigonometry
medium

The value of $\left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right)$ is

A

$0$

B

$1$

C

$2$

D

$\frac{1}{2}$

Solution

 $\tan 1^{\circ} \cdot\tan 2^{\circ}\cdot\tan 3^{\circ} \ldots \tan 89^{\circ}$

$=\tan 1^{\circ}\cdot\tan 2^{\circ}\cdot\tan 3^{\circ} \ldots \tan 44^{\circ} \cdot \tan 45^{\circ} \cdot \tan 46^{\circ} \ldots \tan 87^{\circ}-\tan 88^{\circ} \tan 89^{\circ}$

$=\tan 1^{\circ}\cdot\tan 2^{\circ}\cdot\tan 3^{\circ} \ldots \tan 44^{\circ} \cdot(1)\cdot\tan \left(90^{\circ}-44^{\circ}\right) \ldots \tan \left(90^{\circ}-3^{\circ}\right)$

$\tan \left(90^{\circ}-2^{\circ}\right)\cdot\tan \left(90^{\circ}-1^{\circ}\right)\left(\therefore \tan 45^{\circ}=1\right)$

$=\tan 1^{\circ}\cdot\tan 2^{\circ}\cdot\tan 3^{\circ} \ldots . \tan 44^{\circ}(1) \cdot \cot 44^{\circ} \ldots \ldots \cdot \cot 3^{\circ}-\cot 2^{\circ}-\cot 1^{\circ}$ $\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$

$=\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \ldots \tan 44^{\circ}(1) \cdot \frac{1}{\tan 44^{\circ}} \cdots \frac{1}{\tan 30^{\circ}} \cdot \frac{1}{\tan 2^{\circ}} \cdot \frac{1}{\tan 1^{\circ}}\left[\because \cot \theta=\frac{1}{\tan \theta}\right]$

$=1$

Standard 10
Mathematics

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