Question

Show that \[{\sin ^{ - 1}}\dfrac{{12}}{{13}} + {\cos ^{ - 1}}\dfrac{4}{5} + {\tan ^{ - 1}}\dfrac{{63}}{{16}} = \pi \]

Answer 1

Hint: Here we will convert all the quantities in terms of \[{\tan ^{ - 1}}\theta \] first using the trigonometric ratios and then use the following identity to get the desired solution.
\[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\]
We will also make use of the pythagoras theorem in the solution.

Complete step-by-step answer:
Let us first consider the left hand side:
 \[LHS = {\sin ^{ - 1}}\dfrac{{12}}{{13}} + {\cos ^{ - 1}}\dfrac{4}{5} + {\tan ^{ - 1}}\dfrac{{63}}{{16}}\]………………………. (1)
We know that,
\[\theta = {\sin ^{ - 1}}\left( {\dfrac{{perpendicular}}{{hypotenuse}}} \right)\]
Hence, \[perpendicular = 12\]
\[hypotenuse = 13\]
Now according to Pythagoras theorem we know that,
\[{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {perpendicular} \right)^2}\]
Putting in the respective values we get:-
\[{\left( {13} \right)^2} = {\left( {base} \right)^2} + {\left( {12} \right)^2}\]
Simplifying it further we get:-
\[{\left( {base} \right)^2} = 169 - 144\]
\[ \Rightarrow {\left( {base} \right)^2} = 25\]
Taking square root we get:-
\[base = 5\]
Now we know that,
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{{perpendicular}}{{base}}} \right)\]
Putting the values we get:-
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{{12}}{5}} \right)\]…………………….. (2)
Now we know that,
\[\alpha = {\cos ^{ - 1}}\left( {\dfrac{{base}}{{hypotenuse}}} \right)\]
Therefore,
\[base = 4\]
\[hypotenuse = 5\]
Now according to Pythagoras theorem we know that,
\[{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {perpendicular} \right)^2}\]
Putting in the respective values we get:-
\[{\left( 5 \right)^2} = {\left( 4 \right)^2} + {\left( {perpendicular} \right)^2}\]
Simplifying it further we get:-
\[{\left( {perpendicular} \right)^2} = 25 - 16\]
\[ \Rightarrow {\left( {perpendicular} \right)^2} = 9\]
Taking square root we get:-
\[perpendicular = 3\]
Now we know that,
\[\alpha = {\tan ^{ - 1}}\left( {\dfrac{{perpendicular}}{{base}}} \right)\]
Putting the values we get:-
\[\alpha = {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right)\]……………………………. (3)
Now putting the values from equation 2 and equation 3 in equation 1 we get:-
\[LHS = {\tan ^{ - 1}}\left( {\dfrac{{12}}{5}} \right) + {\tan ^{ - 1}}\left( {\dfrac{3}{4}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}\]
Now applying the following identity for first two terms :-
\[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\]
We get:-
\[LHS = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{12}}{5} + \dfrac{3}{4}}}{{1 - \left( {\dfrac{{12}}{5}} \right)\left( {\dfrac{3}{4}} \right)}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}\]
Taking the LCM we get:-
\[LHS = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{12\left( 4 \right) + 3\left( 5 \right)}}{{20}}}}{{1 - \dfrac{{36}}{{20}}}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}\]
Again taking the LCM in denominator we get:-
\[LHS = {\tan ^{ - 1}}\left( {\dfrac{{\dfrac{{48 + 15}}{{20}}}}{{\dfrac{{20 - 36}}{{20}}}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}\]
Simplifying it further we get:-
\[LHS = {\tan ^{ - 1}}\left( {\dfrac{{63}}{{ - 16}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}\]
This implies,
\[LHS = {\tan ^{ - 1}}\left( { - \dfrac{{63}}{{16}}} \right) + {\tan ^{ - 1}}\dfrac{{63}}{{16}}\]
Again applying the following identity:-
\[{\tan ^{ - 1}}A + {\tan ^{ - 1}}B = {\tan ^{ - 1}}\left( {\dfrac{{A + B}}{{1 - AB}}} \right)\]
We get:-
\[LHS = {\tan ^{ - 1}}\left( {\dfrac{{ - \dfrac{{63}}{{16}} + \dfrac{{63}}{{16}}}}{{1 - \left( { - \dfrac{{63}}{{16}}} \right)\left( {\dfrac{{63}}{{16}}} \right)}}} \right)\]
Simplifying it we get:-
\[LHS = {\tan ^{ - 1}}\left( {\dfrac{0}{{1 - \left( { - 1} \right)}}} \right)\]
\[ \Rightarrow LHS = {\tan ^{ - 1}}\left( {\dfrac{0}{2}} \right)\]
\[ \Rightarrow LHS = {\tan ^{ - 1}}\left( 0 \right)\]
Now we know that,
\[\tan \pi = 0\]
\[ \Rightarrow {\tan ^{ - 1}}0 = \pi \]
Hence, substituting the value we get:-
\[ \Rightarrow LHS = \pi \]
Now considering RHS we get:-
\[RHS = \pi \]
Hence, \[LHS = RHS\]
Hence, proved.

Note: Students should use the trigonometric ratios carefully and convert them using the Pythagoras theorem.
In a right angled triangle,
\[{\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {perpendicular} \right)^2}\]