Find the value of tan lº tan 2° tan 3° ... tan 80°
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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 1°*cot 1°* tan 2°*cot 2°…..tan 44° * cot 44° ]
= We know that tanA *cotA=1
Hence, the equation becomes
= 1 x 1 x 1 x 1 x … x 1
= 1 { As 1ⁿ = 1}
Answer
tan 1° tan 2° tan 3° … tan 89° = 1
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