Prove that \[{\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{33}}{{65}}} \right)\]

Hint: To prove the statement at first we have to assume each term in LHS as separate variables. For example consider \[{\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right)\] be equal to a variable x and \[{\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right)\] be equal to another variable y. Then By using the property of trigonometric inverse function we will find out \[\cos x\] and \[\cos y\]. Then we will find corresponding values of \[\sin x\] and \[\sin y\]. Finally we have to find out the value of \[\cos (x + y)\] by substituting the values of \[\cos x\],\[\cos y\],\[\sin x\] and \[\sin y\] obtained earlier and by using the properties of trigonometric inverse function we can prove the given statement.

Complete step by step answer:
The LHS of the statement is given by
\[LHS = {\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right)\] ………………………… (1)
Let \[{\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) = x\] ………………………… (2)
And \[{\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = y\] ………………………… (3)
But we know the property of the trigonometric function that if \[{\cos ^{ - 1}}\theta = A\] then \[\theta = \cos A\]. Hence using this property we can write eq. (2) and (3) as
\[\cos x = \dfrac{4}{5}\] ………………………… (4)
And \[\cos y = \dfrac{{12}}{{13}}\] ………………………… (5)
Again we know the sine and cosine functions are related by
\[\sin \theta = \sqrt {1 - {{\cos }^2}\theta } \] ………………………… (6)
Applying these formulae to eq. (4) and (5), we will obtain,
\[
  \sin x = \sqrt {1 - {{\left( {\dfrac{4}{5}} \right)}^2}} \\
   = \sqrt {\dfrac{9}{{25}}} \\
   = \dfrac{3}{5} \\
\]
                                ………………………………………… (7)
And
 \[
  \sin y = \sqrt {1 - {{\left( {\dfrac{{12}}{{13}}} \right)}^2}} \\
   = \sqrt {\dfrac{{25}}{{169}}} \\
   = \dfrac{5}{{13}} \\
\]
                            ……………………………………………. (8)
We know the formulae that
\[\cos (x + y) = \cos x\cos y - \sin x\sin y\] ……………………………. (9)
Now substituting the values of eq. (2), (3), (4), (5), (6) and (7) in eq. (9) we will get,
\[
  \cos \left[ {{{\cos }^{ - 1}}\left( {\dfrac{4}{5}} \right) + {{\cos }^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right)} \right] = \left( {\dfrac{4}{5}} \right)\left( {\dfrac{{12}}{{13}}} \right) - \left( {\dfrac{3}{5}} \right)\left( {\dfrac{5}{{13}}} \right) \\
   = \dfrac{{48}}{{65}} - \dfrac{3}{{13}} \\
   = \dfrac{{33}}{{65}} \\
\]
               ……………………………….. (10)
Using the property of inverse trigonometric function we can write eq. (10) as
\[{\cos ^{ - 1}}\left( {\dfrac{4}{5}} \right) + {\cos ^{ - 1}}\left( {\dfrac{{12}}{{13}}} \right) = {\cos ^{ - 1}}\left( {\dfrac{{33}}{{65}}} \right)\]
Now the statement is proved.

Note: In alternative method we can apply direct formula given by, \[{\cos ^{ - 1}}A + {\cos ^{ - 1}}B = {\cos ^{ - 1}}\left[ {AB - \sqrt {\left( {1 - {A^2}} \right)\left( {1 - {B^2}} \right)} } \right]\] to prove the statement.