Question

tan1° tan2° -------------tan89° = ?
​

Answer 1

Good Afternoon

Have a bangtan day

Answer 2

$\huge\mathtt\blue{Answer}$

tan 1° tan 2° tan 3° … tan 89°

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}

Answer

tan 1° tan 2° tan 3° … tan 89° = 1