Question

the value of (tan1. tan2. tan3.......... tan89) is​

Answer 1

Answer:

it's 1

Explanation:

coz ..

tan 1 and tan 89 cancel .....tan 2 and tan 88 cancel.... and so on

tan×tan(90-1)

= tan1×cot1

= tan1/tan1

= 1

and so on till we reach tan 45

which is equal to 1.

HOPE DIS HELPS U !!! FOLLOW ME ..NEW ACCOUNT SO ^_°

Answer 2

tan 1° tan 2° tan 3° … tan 89°

= [tan 1° tan 2° … tan 44°] tan 45°[tan (90° – 44°) tan (90° – 43°)… tan (90° – 1°)]

= [tan 1° tan 2° … tan 44°] [cot 44° cot 43°……. cot 1°] × [tan 45°]

= [(tan 1°× cot 1°) (tan 2°× cot 2°)…..(tan 44°× cot 44°)] × [tan 45°]

We know that

tanA × cotA =1

and

tan45° = 1

Hence, the equation becomes

= 1 × 1 × 1 × 1 × …× 1

= 1   {As 1ⁿ = 1}

Therefore, we conclude the answer as tan 1° tan 2° tan 3° … tan 89° = 1