Question

If \[A = {35^ \circ }\], \[B = {15^ \circ }\] and \[C = {40^ \circ }\], then \[\tan A\tan B + \tan B\tan C + \tan C\tan A\] is equal to
A. 0
B. 1
C. 2
D. 3

Answer 1

Hint: Here in this question, we have to find the exact value of a given trigonometric expression by using the tangent sum identity. First rewrite the given angle in the form of addition of two angles, then the standard trigonometric formula tangent sum identity defined as i.e., \[\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x \cdot \tan y}}\] and further simplify by using a definitions and table of standard angles of trigonometric to get the required value.

Complete step by step answer:
Consider the given angles
\[A = {35^ \circ }\], \[B = {15^ \circ }\] and \[C = {40^ \circ }\]
On adding all the angles, we have
\[ \Rightarrow \,\,\,A + B + C = {35^ \circ } + {15^ \circ } + {40^ \circ }\]
\[ \Rightarrow \,\,\,A + B + C = {90^ \circ }\]
Taking ‘tan’ on both sides, then we have
\[ \Rightarrow \,\,\,\tan \left( {A + B + C} \right) = \tan \left( {{{90}^ \circ }} \right)\] ------(1)
By the definition of trigonometric ratios, the tangent is the ratio of cosine and sine ratios i.e., \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
Then RHS of equation (1) becomes
\[ \Rightarrow \,\,\,\tan \left( {A + B + C} \right) = \dfrac{{\sin \left( {{{90}^ \circ }} \right)}}{{\cos \left( {{{90}^ \circ }} \right)}}\] ------(2)
Now apply a trigonometric tangent sum identity in LHS part i.e., \[\tan \left( {x + y} \right) = \dfrac{{\tan x + \tan y}}{{1 - \tan x \cdot \tan y}}\].
In equation (2), \[x = A + B\] and \[y = C\], then
\[ \Rightarrow \,\,\,\dfrac{{\tan \left( {A + B} \right) + \tan \left( C \right)}}{{1 - \tan \left( {A + B} \right) \cdot \tan \left( C \right)}} = \dfrac{{\sin \left( {{{90}^ \circ }} \right)}}{{\cos \left( {{{90}^ \circ }} \right)}}\] -----(3)
We know the table of standard angle of trigonometric ratios, the value of \[\sin \left( {{{90}^ \circ }} \right) = 1\] and \[\cos \left( {{{90}^ \circ }} \right) = 1\], on substituting the values in equation (3) we have
 \[ \Rightarrow \,\,\,\dfrac{{\tan \left( {A + B} \right) + \tan \left( C \right)}}{{1 - \tan \left( {A + B} \right) \cdot \tan \left( C \right)}} = \dfrac{1}{0}\]
On equating the denominators of LHS and RHS, then
\[ \Rightarrow \,\,\,1 - \tan \left( {A + B} \right) \cdot \tan \left( C \right) = 0\]
Subtract 1 on both the sides, then
\[ \Rightarrow \,\,\, - \tan \left( {A + B} \right) \cdot \tan \left( C \right) = - 1\]
Cancelling the ‘\[ - \,\,ve\]‘ sign on both sides, we get
\[ \Rightarrow \,\,\,\tan \left( {A + B} \right) \cdot \tan \left( C \right) = 1\]
Again, by the sum identity of tangent ratio
\[ \Rightarrow \,\,\,\left[ {\dfrac{{\tan A + \tan B}}{{1 - \tan A \cdot \tan B}}} \right] \cdot \tan \left( C \right) = 1\]
Multiply both side by \[1 - \tan A \cdot \tan B\], then
\[ \Rightarrow \,\,\,\left( {\tan A + \tan B} \right) \cdot \tan C = 1 - \tan A \cdot \tan B\]
\[ \Rightarrow \,\,\,\tan A \cdot \tan C + \tan B \cdot \tan C = 1 - \tan A \cdot \tan B\]
Add ‘\[\tan A \cdot \tan B\]’ on both side, we get
\[ \Rightarrow \,\,\,\tan A \cdot \tan C + \tan B \cdot \tan C + \tan A \cdot \tan B = 1\]
Or
\[\therefore \,\,\,\,\,\,\tan A \cdot \tan B + \tan B \cdot \tan C + \tan C \cdot \tan A = 1\].
Hence, the required value is 1.

So, the correct answer is “Option B”.

Note: When the question is based on trigonometric function, we must know about the definitions of all six trigonometric ratios and value of their standard angles. Remember all the basic formulas like trigonometric identity, half and double angle formula, addition and difference identity of trigonometric function and transformation formulas.