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NCERT Exemplar · Q34

Q.The value of tan⁡1∘ tan⁡2∘ tan⁡3∘…tan⁡89∘\tan 1^\circ\,\tan 2^\circ\,\tan 3^\circ \ldots \tan 89^\circ is
(A) 00
(B) 11
(C) 12\dfrac{1}{2}
(D) Not defined

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Pairing complementary angles exploits the identity tan⁡θ⋅tan⁡(90°−θ)=1\tan \theta \cdot \tan(90° - \theta) = 1, collapsing the entire product to tan⁡45°=1\tan 45° = 1.

The key insight here is recognizing the symmetry hidden in this long product. When you multiply tangents of all angles from 1°1° to 89°89°, you're actually multiplying pairs of complementary angles—angles that sum to 90°90°.

Why complementary angles matter

Recall the co-function identity:

tan⁡(90°−θ)=cot⁡θ=1tan⁡θ\tan(90° - \theta) = \cot \theta = \frac{1}{\tan \theta}

This means that for any angle θ\theta, the product tan⁡θ⋅tan⁡(90°−θ)=1\tan \theta \cdot \tan(90° - \theta) = 1. This is the engine that drives our solution.

Step-by-step pairing

  1. Rewrite the product by grouping complementary pairs:

tan⁡1°⋅tan⁡2°⋯tan⁡89°=(tan⁡1°⋅tan⁡89°)⋅(tan⁡2°⋅tan⁡88°)⋯(tan⁡44°⋅tan⁡46°)⋅tan⁡45°\tan 1° \cdot \tan 2° \cdots \tan 89° = (\tan 1° \cdot \tan 89°) \cdot (\tan 2° \cdot \tan 88°) \cdots (\tan 44° \cdot \tan 46°) \cdot \tan 45°

  1. Apply the complementary identity to each pair:

    • tan⁡1°⋅tan⁡89°=tan⁡1°⋅tan⁡(90°−1°)=tan⁡1°⋅cot⁡1°=1\tan 1° \cdot \tan 89° = \tan 1° \cdot \tan(90° - 1°) = \tan 1° \cdot \cot 1° = 1
    • tan⁡2°⋅tan⁡88°=tan⁡2°⋅tan⁡(90°−2°)=tan⁡2°⋅cot⁡2°=1\tan 2° \cdot \tan 88° = \tan 2° \cdot \tan(90° - 2°) = \tan 2° \cdot \cot 2° = 1
    • tan⁡3°⋅tan⁡87°=1\tan 3° \cdot \tan 87° = 1

    And so on, all the way up to:

    • tan⁡44°⋅tan⁡46°=1\tan 44° \cdot \tan 46° = 1
  2. Count the pairs:

    From 1°1° to 89°89°, we have 8989 angles. Pairing them as (1°,89°),(2°,88°),…,(44°,46°)(1°, 89°), (2°, 88°), \ldots, (44°, 46°) gives us 4444 pairs, with 45°45° left unpaired in the middle.

  3. Evaluate the product:

    Each of the 4444 pairs contributes a factor of 11: …

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