Question

\(\tan 1^{\circ} \cdot \tan 2^{\circ} \cdot \tan 3^{\circ} \cdots \tan 89^{\circ} =\)

• A. 2
• B. 0
• C. 3
• D. 1

Hint:

Whenever a product contains \(\tan\theta\) and \(\tan(90^\circ-\theta)\), pair them immediately because their product equals 1.

Answer 1

The Correct Option is D

Concept: This problem is based on the complementary angle identity of trigonometry: \[ \tan \theta=\cot(90^\circ-\theta) \] Since \[ \cot \alpha=\frac{1}{\tan \alpha} \] we obtain \[ \tan \theta \cdot \tan(90^\circ-\theta)=1 \] This property helps simplify long products involving tangent functions.

Step 1: Pair complementary angles.
Given \[ P=\tan1^\circ \tan2^\circ \tan3^\circ \cdots \tan89^\circ \] Pair the first and last terms: \[ (\tan1^\circ \tan89^\circ) (\tan2^\circ \tan88^\circ) (\tan3^\circ \tan87^\circ) \cdots (\tan44^\circ \tan46^\circ) \tan45^\circ \] Now, \[ \tan1^\circ \tan89^\circ =\tan1^\circ \cot1^\circ =1 \] Similarly, \[ \tan2^\circ \tan88^\circ=1 \] and every complementary pair gives 1.

Step 2: Evaluate the middle term.
The only unpaired term is \[ \tan45^\circ=1 \] Therefore, \[ P=1\times1\times1\times\cdots\times1\times1 \] \[ P=1 \] 1