Question

In a $\Delta ABC$ , if $\cos A\cos B\cos C = \dfrac{{\sqrt 3 - 1}}{8}$ and $\sin A\sin B\sin C = \dfrac{{3 + \sqrt 3 }}{8}$ , then the value of $\tan A\tan B + \tan B\tan C + \tan C\tan A$ is

Answer 1

Hint: To find the value, we convert all the terms of the given equation into the form of sin and cos functions using the definition of $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$, such that the given values of sin and cos functions can be used.

Complete step-by-step answer:
Given Data,
$\cos A\cos B\cos C = \dfrac{{\sqrt 3 - 1}}{8}$ and $\sin A\sin B\sin C = \dfrac{{3 + \sqrt 3 }}{8}$ .
Now, we have to find the value of $\tan A\tan B + \tan B\tan C + \tan C\tan A$
As we know the value of tangent of an angle in terms of sine and cosine is given as:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
So, let us substitute the value in the above function.
$
   = \tan A\tan B + \tan B\tan C + \tan C\tan A \\
   = \dfrac{{\sin A}}{{\cos A}}\dfrac{{\sin B}}{{\cos B}} + \dfrac{{\sin B}}{{\cos B}}\dfrac{{\sin C}}{{\cos C}} + \dfrac{{\sin C}}{{\cos C}}\dfrac{{\sin A}}{{\cos A}} \\
 $
Let us now take the LCM to solve further.
  $ = \dfrac{{\sin A\sin B\cos C + \sin C\sin B\cos A + \sin A\sin C\cos B}}{{\cos A\cos B\cos C}}$
Now let us take some term common in the numerator to solve further by the use of formula.
$ = \dfrac{{\sin A\sin B\cos C + \sin C\left( {\sin B\cos A + \sin A\cos B} \right)}}{{\cos A\cos B\cos C}}$
As we know that the formula for sum of the sine of the angle is given as:
$ \Rightarrow \sin \left( {x + y} \right) = \sin x\cos y + \sin y\cos x$
Let us use the above formula in the above term to solve further.
$ = \dfrac{{\sin A\sin B\cos C + \sin C\sin \left( {A + B} \right)}}{{\cos A\cos B\cos C}}$
Given that angle A, B and C are part of the triangle so we have the relation between the angles of the triangle given as:
$
   \Rightarrow A + B + C = {180^0} \\
   \Rightarrow A + B = {180^0} - C \\
   \Rightarrow \sin \left( {A + B} \right) = \sin \left( {{{180}^0} - C} \right) \\
   \Rightarrow \sin \left( {A + B} \right) = \sin C - - - - \left( 1 \right){\text{ }}\left[ {\because \sin \left( {{{180}^0} - \theta } \right) = \sin \theta } \right] \\
   \Rightarrow \cos \left( {A + B} \right) = \cos \left( {{{180}^0} - C} \right) \\
   \Rightarrow \cos \left( {A + B} \right) = - \cos C - - - - \left( 2 \right){\text{ }}\left[ {\because \cos \left( {{{180}^0} - \theta } \right) = \cos \theta } \right] \\
 $
Let us substitute this result from equation (1) in the given term.
$
   = \dfrac{{\sin A\sin B\cos C + \sin C\sin C}}{{\cos A\cos B\cos C}}\;{\text{ }}\left[ {\because \sin \left( {A + B} \right) = \sin C} \right] \\
   = \dfrac{{\sin A\sin B\cos C + {{\sin }^2}C}}{{\cos A\cos B\cos C}} \\
 $
We know the identity
$
  {\sin ^2}\theta + {\cos ^2}\theta = 1 \\
   \Rightarrow {\sin ^2}\theta = 1 - {\cos ^2}\theta \\
 $
Let us substitute this in the above problem and let us take some common terms in the numerator.
$
   = \dfrac{{\sin A\sin B\cos C + 1 - {{\cos }^2}C}}{{\cos A\cos B\cos C}} \\
   = \dfrac{{1 + \cos C\left( { - \cos C + \sin A\sin B} \right)}}{{\cos A\cos B\cos C}} \\
 $
Let us substitute the value from equation (2) in the above equation.
\[
   = \dfrac{{1 + \cos C\left( {\cos \left( {A + B} \right) + \sin A\sin B} \right)}}{{\cos A\cos B\cos C}}{\text{ }}\left[ {\because \cos \left( {A + B} \right) = - \cos C} \right] \\
   = \dfrac{{1 + \cos C\left( {\cos A\cos B - \sin A\sin B + \sin A\sin B} \right)}}{{\cos A\cos B\cos C}}{\text{ }}\left[ {\because \cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B} \right] \\
   = \dfrac{{1 + \cos C\left( {\cos A\cos B} \right)}}{{\cos A\cos B\cos C}} \\
   = \dfrac{1}{{\cos A\cos B\cos C}} + 1 \\
 \]
Now let us substitute the value of terms given in the problem to find the final value.
\[ \Rightarrow \dfrac{1}{{\cos A\cos B\cos C}} + 1 = \dfrac{1}{{\left( {\dfrac{{\sqrt 3 - 1}}{8}} \right)}} + 1{\text{ }}\left[ {\because \cos A\cos B\cos C = \dfrac{{\sqrt 3 - 1}}{8}} \right]\]
Let us now simplify the term.
\[
   = \dfrac{{8 + \sqrt 3 - 1}}{{\sqrt 3 - 1}} \\
   = \dfrac{{7 + \sqrt 3 }}{{\sqrt 3 - 1}} \\
   \Rightarrow \tan A\tan B + \tan B\tan C + \tan C\tan A = \dfrac{{7 + \sqrt 3 }}{{\sqrt 3 - 1}} \\
 \]
Hence, the value of $\tan A\tan B + \tan B\tan C + \tan C\tan A$ is \[\dfrac{{7 + \sqrt 3 }}{{\sqrt 3 - 1}}\]

Note: In order to solve problems of this type the key is to observe that as the given angles are of a triangle $\sin \left( {A + B} \right) = \sin C$ , as the sum of angles in a triangle is 180 degrees. Adequate knowledge in the trigonometric conversions and formulae is required to re-arrange the given equation and simplify.