The value of (tan1° tan2° tan3° ... tan89°) is
a. 0
b. 1
c. 2
d. 1/2

Solution:

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

Using the trigonometric ratios of complementary angles,

tan (90° - A) = cot A

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.

✦ Try This: The value of (cot1° cot 2° cot3° ... cot89°) is

We have to find the value of cot1° cot 2° cot3° ... cot89°

Using the trigonometric ratios of complementary angles,

cot (90° - A) = tan A

The expression can be written as

(cot1° cot2° cot3° ... cot 44°) × cot 45° × (cot(90°-44°) cot(90°- 43°) cot(90°- 42°) ... cot(90°-1°))

= (cot1° cot2° cot3° ... cot44°) × cot45° × (tan44° tan43° tan42°……..tan1°)

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 (cot1° cot2° cot3° ... cot89°) is 1.

☛ Also Check: NCERT Solutions for Class 10 Maths Chapter 8

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NCERT Exemplar Class 10 Maths Exercise 8.1 Problem 6

The value of (tan1° tan2° tan3° ... tan89°) is a. 0, b. 1, c. 2, d. 1/2

Summary:

The value of (tan1° tan2° tan3° ... tan89°) is 1

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