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Q.

Evaluate : (tan1° tan2° tan3° ... tan89°)

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a

2

b

1

c

0

d

3

answer is B.

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Detailed Solution

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We have to find the value of (tan1° tan2° tan3° ... tan89°)

Using the trigonometric ratios of complementary angles, we get:

tan (90° - A) = cot A

So, The expression can be written as:

= (tan1° tan2° tan3° ... tan 44°) × tan 45° × (tan(90°- 44°) tan(90°- 43°) tan (90°- 42°) ... tan(90°- 1°))

= (tan1° tan2° tan3° ... tan 44°) × tan 45° × (cot44° cot 43° cot42°……..cot1°)

We know that tan A × cot A = 1

So, (tan1° × cot1°) × (tan2°× cot2°) × (tan3° × cot3°) ×…...tan 45° ×……….(tan44° × cot44°)

= 1 × 1 × 1 × ……………..× 1

= 1

Therefore, the value of (tan1° tan2° tan3° ... tan89°) is 1.

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Evaluate : (tan1° tan2° tan3° ... tan89°)