Question

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

Answer 1

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