# 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) $\frac{1}{2}$

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Given:

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

To do:

We have to find the value of $\left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right)$.

Solution:

We know that,

$tan\ (90^{\circ}- \theta) = cot\ \theta$

$tan\ \theta \times \cot\ \theta=1$

Therefore,

$\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ}.....\tan 45^{\circ}...... \tan 87^{\circ} \tan 88^{\circ} \tan 89^{\circ}=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ}......\tan 45^{\circ}.........\tan (90^{\circ}-3^{\circ}) \tan (90^{\circ}-2^{\circ}) \tan (90^{\circ}-1^{\circ})$

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

$=(\tan 1^{\circ} \cot 1^{\circ})(\tan 2^{\circ} \cot 2^{\circ})..................(\tan 44^{\circ} \cot 44^{\circ})(1)$

$=1$

Updated on 10-Oct-2022 13:28:52