Geometric mean of $\tan 1^{\circ}, \tan 2^{\circ}, \ldots \ldots . . .$, $\tan 89^{\circ}$ is

Question

Aicharya · Accepted Answer

$\tan 1^{\circ} \cdot \tan 2^{\circ} \ldots \tan 45^{\circ} \cdot \tan 46^{\circ} \ldots \tan 88^{\circ} \cdot \tan 89^{\circ}$ $\tan 1^{\circ} \cdot \tan 2^{\circ} \ldots \tan 44^{\circ} \cdot \tan 45^{\circ} \tan \left(90-44^{\circ}\right)$ $\ldots \tan \left(90^{\circ}-1^{\circ}\right)$ $\tan 1^{\circ} \cdot \tan 2^{\circ} \ldots \tan 44^{\circ} \tan 45^{\circ} \operatorname{Cot} 44^{\circ} \cdot \operatorname{Cot} 43^{\circ}$... $\operatorname{Cot} 1^{\circ}$ $\tan 1^{\circ} \cdot \tan 2^{\circ} \ldots \tan 44^{\circ} 1 \frac{1}{\tan 44^{\circ}} \frac{1}{\tan 43^{\circ}} \cdots \frac{1}{\tan 1^{\circ}}=1$ Hence, option (b) is correct.

Geometric mean of \(\tan 1^{\circ}, \tan 2^{\circ}, \ldots \ldots . . .\), \(\tan 89^{\circ}\) is

Solution