Question

The value of $tan\;27\;tan\;31 + tan\;32\;tan\;31 + tan\;31\;tan\;27\;$ is
A.0
B.1
C.2
D.Can not be determined

Answer 1

Hint : To solve this type of question try to convert in standard forms as we do not know the value of $tan\;31$ or $tan\;32$ .Let the angles be A,B and C. So to convert in standard form just add all the given angles and find the angle C then find the $tan\;C$ in terms of $\tan B$ and $\tan B$ . After finding all these things just put the value of $tan\;C$ and simplify then we will get our answer.

Complete step-by-step answer :
From given, we have,
Suppose we add all the given angles
So we have
$27 + 32 + 31 = 90$
And suppose the given trigonometric expression as S
$S = tan\;A\;tan\;B + tan\;B\;tan\;C + tan\;C\;tan\;A$ .......(1)
Or after taking $tan\;C$ common we get,
$\Rightarrow S = tan\;A\;tan\;B + tan\;C\left( {tan\;A + tan\;B} \right)$
where $A + B + C = 90$ , and hence, $C = 90 - \left( {A + B} \right)$
now taking tan both side
so we have
$tan\;C = tan\;\left( {90 - A - B} \right)$
or,
$tan\;C = cot\;\left( {A + B} \right)$
or,
$tan\;C{\text{ }} = \dfrac{1}{{tan\;\left( {A + B} \right)}}$
now applying the $tan\;\left( {A + B} \right)$ formula we get
$tan\;C = \dfrac{{1 - tan\;A\;tan\;B}}{{tan\;A + tan\;B}}$
on cross multiplying
$tan\;C\;\left( {tan\;A + tan\;B} \right) = 1 - tan\;A\;tan\;B$ ....(2)
from (1) and (2)
we have the value of S
$S = tan\;A\;tan\;B + tan\;C\;\left( {tan\;A + tan\;B} \right)$
$= tan\;A\;\;tan\;B + \left( {1 - tan\;A\;\;tan\;B} \right)$
Here $tan\;A\;tan\;B$ terms will cancel out each other.
So we have S is equal to
$= 1$
Hence the value of the expression
$tan\;27\;tan\;31 + tan\;32\;tan\;31 + tan\;31\;tan\;27\;$ = $1$
So, the correct answer is “Option B”.

Note : To solve this type of question we can use the formula of $tan\;\left( {A + B + C} \right)$ rather than $tan\;\left( {A + B} \right)$ but in this way may you have to face quite more difficulties than the method used in this solution.