Question

6. The value of (tanlº tan2° tan3º ... tan89) is
(C) 2
1
(D)
(B)
0
(A)
2​

Answer 1

Answer :

1

Solution :

• To find : tan1°×tan2°×tan3°× . . . ×tan89°

• Note : tan(90° - ∅) = cot∅

tan∅×cot∅ = 1

Now ,

Let the given expression be x .

Thus ,

→ x = tan1° × tan2° × tan3° × . . . × tan89°

→ x = tan1° × tan2° × tan3° × . . . × tan45° ×

. . . × tan89°

Now ,

Rearranging with the complementary angles , we get ;

→ x = ( tan1° × tan89° ) × ( tan2° × tan88° ) × ( tan3°×tan87° ) × . . . × tan45°

→ x = ( tan1°×tan(90°-1°) ) × ( tan2°×tan(90°-2°) ) × ( tan3°×tan(90°-3°) ) × . . . × tan45°

→ x = ( tan1°×cot1° ) × ( tan2°×cot2° ) × ( tan3°×cot3° ) × . . . × tan45°

→ x = 1 × 1 × 1 × . . . × 1

→ x = 1

Hence ,

tan1° × tan2° × tan3° × . . . × tan89° = 1