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.