# Find the value of $\tan {1^ \circ }\tan {2^ \circ }\tan {3^ \circ }.......\tan {89^ \circ }$

from the options given below

A. 0

B. 1

C. 2

D. 3

Last updated date: 30th Mar 2023

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Answer

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Hint-Here, let us try to solve this question by making use of the formula

$\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta $ and solve

By making use of the formula $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta $,we

can write

$\tan {89^ \circ }$= $\tan ({90^ \circ } - {1^ \circ }) = \cot {1^ \circ }$

Similarly we can write

tan${88^ \circ }$ =$\tan ({90^ \circ } - {2^ \circ }) = \cot {2^ \circ }$

On proceeding in a similar manner we can write the value of tan in terms of cot upto

$\tan {46^ \circ }$ and the value of $\tan {45^ \circ }$ is retained as it is and not converted to

cot. This is

because if we pair $\tan {89^ \circ }$,$\tan {1^ \circ }$ ; $\tan {2^ \circ },\tan {88^ \circ }$ ;we can pair them up to $\tan {44^ \circ }\tan {46^ \circ }$

and finally $\tan {45^ \circ }$ will remain unpaired with any other element.

So, now the equation becomes $(\tan {1^ \circ }\cot {1^ \circ })(\tan {2^ \circ }\cot {2^ \circ

}).....(\tan {44^ \circ }\cot {44^ \circ })(\tan {45^ \circ })$

Since tan and cot are reciprocals of each other $(\tan {1^ \circ }\cot {1^ \circ })(\tan {2^ \circ

}\cot {2^ \circ })...$ will cancel out

and will become 1 and the value of $\tan {45^ \circ }$ will also become 1.

So, the equation will now be equal to (1)(1)……..(1)(1)=1

So, therefore the value of $\tan {1^ \circ }\tan {2^ \circ }\tan {3^ \circ }.......\tan {89^ \circ

}$=1 $$ $$

Note: To solve these kind of problems we will make use of the complementary angle formula

of the trigonometric ratios

$\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta $ and solve

By making use of the formula $\tan \left( {{{90}^ \circ } - \theta } \right) = \cot \theta $,we

can write

$\tan {89^ \circ }$= $\tan ({90^ \circ } - {1^ \circ }) = \cot {1^ \circ }$

Similarly we can write

tan${88^ \circ }$ =$\tan ({90^ \circ } - {2^ \circ }) = \cot {2^ \circ }$

On proceeding in a similar manner we can write the value of tan in terms of cot upto

$\tan {46^ \circ }$ and the value of $\tan {45^ \circ }$ is retained as it is and not converted to

cot. This is

because if we pair $\tan {89^ \circ }$,$\tan {1^ \circ }$ ; $\tan {2^ \circ },\tan {88^ \circ }$ ;we can pair them up to $\tan {44^ \circ }\tan {46^ \circ }$

and finally $\tan {45^ \circ }$ will remain unpaired with any other element.

So, now the equation becomes $(\tan {1^ \circ }\cot {1^ \circ })(\tan {2^ \circ }\cot {2^ \circ

}).....(\tan {44^ \circ }\cot {44^ \circ })(\tan {45^ \circ })$

Since tan and cot are reciprocals of each other $(\tan {1^ \circ }\cot {1^ \circ })(\tan {2^ \circ

}\cot {2^ \circ })...$ will cancel out

and will become 1 and the value of $\tan {45^ \circ }$ will also become 1.

So, the equation will now be equal to (1)(1)……..(1)(1)=1

So, therefore the value of $\tan {1^ \circ }\tan {2^ \circ }\tan {3^ \circ }.......\tan {89^ \circ

}$=1 $$ $$

Note: To solve these kind of problems we will make use of the complementary angle formula

of the trigonometric ratios

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