Find the value tan 1-tan 2-tan 3- - tan 89-

The correct option is C 1tan1^∘. tan2^∘.tan3^∘.... ......tan44^∘. tan45^∘.tan46^∘.........tan89^∘ = tan1^∘.tan2^∘.tan3^∘.....tan44^∘.1.cot(90^∘ 46^∘).. ..co ...

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Byju's Answer

Standard X

Mathematics

Complementary Trigonometric Ratios

Find the valu...

Question

Find the value
$t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . . . . . . . . t a n 89^{\circ}$

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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
$t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . . . . . . . . t a n 44^{\circ} . t a n 45^{\circ} . t a n 46^{\circ} . . . . . . . . . t a n 89^{\circ} = t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . . . t a n 44^{\circ} .1 . c o t (90^{\circ} - 46^{\circ}) . . . . c o t (90^{\circ} - 89^{\circ}) = t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . . . . . . . t a n 44^{\circ} c o t 44^{\circ} . . . . . . . . c o t 2^{\circ} c o t 1^{\circ}$
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, $t a n 1^{\circ} . c o t 1^{\circ} . t a n 2^{\circ} c o t 2^{\circ} . t a n 2^{\circ} . c o t 3^{\circ} . . . . . . . . t a n 44^{\circ} . c o t 44^{\circ} = 1$

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Similar questions

Q. Find the value of

t a n 1^{\circ} t a n 2^{\circ} t a n 3^{\circ}, . . . . . . . . . . . . . . . . t a n 89^{\circ}

___

Q. The value of

tan 1^{\circ} tan 2^{\circ} tan 3^{\circ} \dots tan 89^{\circ}

The value of tan $1^{\circ}$ $\times$ tan $2^{\circ}$ $\times$ tan $3^{\circ}$ $\times$ …... $\times$ tan $89^{\circ}$ is :

Q. The value of

tan 1^{\circ} tan 2^{\circ} tan 3^{\circ} . . . . . tan 89^{\circ}

Q. The value of tan 1° tan 2° tan 3° _________ tan 89° is _________.

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Find the value tan1∘.tan2∘.tan3∘..........tan89∘

Find the value
$t a n 1^{\circ} . t a n 2^{\circ} . t a n 3^{\circ} . . . . . . . . . . t a n 89^{\circ}$