Question

the value of (than 1,than 2,than 3).....(than 89) is ​

Answer 1

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 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}

Final Answer -

tan 1° tan 2° tan 3° … tan 89° = 1