Tan1- -tan2-tan3-tan89-

Given, nbsp;tan1^∘. tan2^∘.tan3^∘.....tan44^∘. tan45^∘.tan46^∘....tan89^∘ [∵ tan θ = cot (90 θ)] ∴ tan1^∘. tan2^∘.tan3^∘.....tan44^∘. tan45^∘.tan46^∘.....tan8 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Variation of Trigonometric Ratios from 0 to 90 Degrees

tan1∘ .tan2∘....

Question

$t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . t a n 89^{\circ}$ =_________.

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

t a n 90^{\circ}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\sqrt{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B 1
Given, $t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . . . t a n 44^{\circ} . t a n 45^{\circ} . t a n 46^{\circ} . . . . t a n 89^{\circ}$
$[∵ t a n θ = c o t (90 - θ)]$
$∴ t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . . . t a n 44^{\circ} . t a n 45^{\circ} . t a n 46^{\circ} . . . . . t a n 89^{\circ} =$ $t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . . . t a n 44^{\circ} . t a n 45^{\circ} . c o t (90^{\circ} - 46^{\circ}) . . . . c o t (90^{\circ} - 89^{\circ})$
$\Rightarrow t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . . . t a n 44^{\circ} c o t 44^{\circ} . . . . c o t 2^{\circ} c o t 1^{\circ}$
Rearranging the terms,
$= t a n 1^{\circ} c o t 1^{\circ} . t a n 2^{\circ} c o t 2^{\circ} . t a n 2^{\circ} . c o t 3^{\circ} . . . . . t a n 44^{\circ} c o t 44^{\circ}$
Now, we know that $t a n θ = \frac{1}{c o t θ}$
$\Rightarrow t a n θ \times c o t θ = 1$
Thus, $t a n 1^{\circ} . c o t 1^{\circ} . t a n 2^{\circ} c o t 2^{\circ} . t a n 2^{\circ} . c o t 3^{\circ} . . . . . t a n 44^{\circ} . c o t 44^{\circ} = 1$

Suggest Corrections

tan1∘.tan2∘.tan3∘...tan89∘=_________.

$t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . t a n 89^{\circ}$ =_________.