Tan 1-tan 2-tan 3-tan 89-

The correct option is B 1Given, nbsp;tan1^∘. tan2^∘.tan3^∘.....tan44^∘. tan45^∘.tan46^∘....tan89^∘ [∵ tan θ = cot (90 θ)] ∴ tan1^∘. tan2^∘.tan3^∘.....tan44^∘ ...

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Byju's Answer

Standard IX

Mathematics

Complementary Ratios

tan 1∘·tan 2∘...

Question

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

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t a n 90^{\circ}

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\sqrt{3}

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Solution

The correct option is A 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∘=_________.

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