14. Find the value of tan 1° tan 2° tan 3° ... tan 80°
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Answer:
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
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