The value of tan1
Question:

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

(a) 0

(b) 1

(c) $\frac{1}{2}$

(d) not defined

Solution:

We know that, $\tan \left(90^{\circ}-\theta\right)=\cot \theta$

So,

$\tan 89^{\circ}=\tan \left(90^{\circ}-1^{\circ}\right)=\cot 1^{\circ}$

$\tan 88^{\circ}=\tan \left(90^{\circ}-2^{\circ}\right)=\cot 2^{\circ}$

$\tan 87^{\circ}=\tan \left(90^{\circ}-3^{\circ}\right)=\cot 3^{\circ}$

$\tan 46^{\circ}=\tan \left(90^{\circ}-44^{\circ}\right)=\cot 44^{\circ}$

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

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

$=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 44^{\circ} \tan 45^{\circ} \cot 44^{\circ} \ldots \cot 3^{\circ} \cot 2^{\circ} \cot 1^{\circ}$

$=\left(\tan 1^{\circ} \cot 1^{\circ}\right)\left(\tan 2^{\circ} \cot 2^{\circ}\right)\left(\tan 3^{\circ} \cot 3^{\circ}\right) \ldots\left(\tan 44^{\circ} \cot 44^{\circ}\right) \tan 45^{\circ}$

$=1 \quad\left(\tan 45^{\circ}=1\right.$ and $\left.\tan \theta \cot \theta=1\right)$

Hence, the correct answer is option B.

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