What is the value of $cot\ 15^ocot\ 16^ocot\ 17^o.....cot\ 73^ocot\ 74^ocot\ 75^o$?

Given:

$cot\ 15^ocot\ 16^ocot\ 17^o.....cot\ 73^ocot\ 74^ocot\ 75^o$.
To do:

We have to find the value of $cot\ 15^ocot\ 16^ocot\ 17^o.....cot\ 73^ocot\ 74^ocot\ 75^o$.
Solution:

We know that,

$cot\ \theta.tan\ \theta=1$

$cot\ \theta=tan\ (90^o-\theta)$

 This implies,

$cot\ (90^-75)^o=tan\ 15^o$

$cot\ (90^-74)^o=tan\ 16^o$

$cot\ (90^-73)^o=tan\ 17^o$

$cot\ (90^-46)^o=tan\ 44^o$

Therefore,

$cot\ 15^ocot\ 16^ocot\ 17^o.....cot\ 73^ocot\ 74^ocot\ 75^o=cot\ 15^ocot\ 16^ocot\ 17^o.....cot\ 44^ocot\ 45^otan\ 44^......tan\ 17^otan\ 16^otan\ 15^o$

$=cot\ 15^otan\ 15^ocot\ 16^otan\ 16^ocot\ 17^otan\ 17^o.......cot\ 45^o$

$=1\times1\times1.........1$

$=1$

Therefore,

$cot\ 15^ocot\ 16^ocot\ 17^o.....cot\ 73^ocot\ 74^ocot\ 75^o=1$.

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Updated on: 10-Oct-2022

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