The value of tan 1o tan 2o tan 3o....tan89o is equal to

Question

The value of tan 1o tan 2o tan 3o....tan89o is equal to (A)  -1  (B)  2  (C)  (π /2)  (D)  1

Tardigrade Admin · Accepted Answer

Given tan 1o tan 2o tan 3o.... tan 89o Now, tan 89o= of tan (90-1)= cot 1o tan 88o= cot 2o and so on. ∴ tan 1o tan 2o..... tan 89o =( tan 1o cot 1o)(tan2o cot 2o).....tan45o 1⋅ 1=1

Q. The value of $tan 1^{o} tan 2^{o} tan 3^{o} .... t an 89^{o}$ is equal to

$- 1$

$2$

$\frac{π}{2}$

$1$

Given $tan 1^{o} tan 2^{o} tan 3^{o} .... tan 89^{o}$
Now, $tan 89^{o} =$ of $tan (90 - 1) = cot 1^{o}$
$tan 88^{o} = cot 2^{o}$ and so on.
$∴$ $tan 1^{o} tan 2^{o} ..... tan 89^{o}$
$= (tan 1^{o} cot 1^{o}) (t an 2^{o} cot 2^{o}) ..... t an 45^{o}$
$1 \cdot 1 = 1$

Q. The value of tan1otan2otan3o....tan89o is equal to

−1

2

2π​

1

Given tan1otan2otan3o....tan89o Now, tan89o= of tan(90−1)=cot1o tan88o=cot2o and so on. ∴ tan1otan2o.....tan89o =(tan1ocot1o)(tan2ocot2o).....tan45o 1⋅1=1

Q. The value of $tan 1^{o} tan 2^{o} tan 3^{o} .... t an 89^{o}$ is equal to

$- 1$

$2$

$\frac{π}{2}$

$1$

Given $tan 1^{o} tan 2^{o} tan 3^{o} .... tan 89^{o}$
Now, $tan 89^{o} =$ of $tan (90 - 1) = cot 1^{o}$
$tan 88^{o} = cot 2^{o}$ and so on.
$∴$ $tan 1^{o} tan 2^{o} ..... tan 89^{o}$
$= (tan 1^{o} cot 1^{o}) (t an 2^{o} cot 2^{o}) ..... t an 45^{o}$
$1 \cdot 1 = 1$