The value of

Question:

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:

(b) tan1°-tan2°-tan3°… tan 89°

= tan1°-tan2°-tan3°… tan44° . tan 45° . tan 46°… tan 87°-tan 88°tan 89°

= tan 1°- tan2 °- tan 3°… tan 44° . (1)- tan (90° – 44°)… tan (90° – 3°)

tan (90° -2°)- tan (90° -1°) (∴ tan 45° = 1)

= tan1°-tan2°-tan3°…. tan44° (1) . cot 44°……. cot3°-cot2°-cot1°

$\left[\because \tan \left(90^{\circ}-\theta\right)=\cot \theta\right]$

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

$=1$

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