Show that the given expression ${{\sin }^{-1}}\left( \dfrac{12}{13} \right)+{{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\tan }^{-1}}\left( \dfrac{63}{16} \right)=\pi $
VerifiedHint: We will use the formulas for trigonometric ratios which is given by $\cos \left( \theta \right)=\dfrac{\text{Base}}{\text{Hypotenuse}}$ and $\sin \left( a \right)=\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$. This will help us to solve the question. And with the help of Pythagoras theorem we will find the angles.
Complete step-by-step answer:
We will consider the expression ${{\sin }^{-1}}\left( \dfrac{12}{13} \right)+{{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\tan }^{-1}}\left( \dfrac{63}{16} \right)=\pi $. First we will substitute ${{\sin }^{-1}}\left( \dfrac{12}{13} \right)=a$. By taking the inverse sine function to the right side of the expression we will have $\left( \dfrac{12}{13} \right)=\sin \left( a \right)$. Now as we know that $\sin \left( a \right)=\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$. Thus we have perpendicular as 12 units and hypotenuse is 13 units. By using Pythagoras theorem here we will find the value of the base. Therefore, we have the following diagram.
Thus by using Pythagoras theorem we have
\[ {{\left( 13 \right)}^{2}}={{\left( 12 \right)}^{2}}+{{x}^{2}} \]
\[ \Rightarrow {{\left( 13 \right)}^{2}}-{{\left( 12 \right)}^{2}}={{x}^{2}} \]
\[ \Rightarrow 169-144={{x}^{2}} \]
\[ \Rightarrow {{x}^{2}}=25 \]
\[ \Rightarrow x=\pm 5 \]
As no side can be a negative side therefore we choose x = 5 units here. Now this will work as a base of the triangle. Therefore by applying the formula $\cos \left( \theta \right)=\dfrac{\text{Base}}{\text{Hypotenuse}}$ we have $\cos \left( a \right)=\dfrac{5}{13}$.
As we know that $\tan \left( a \right)=\dfrac{\sin \left( a \right)}{\cos \left( a \right)}$. Therefore we have
\[ \tan \left( a \right)=\dfrac{\dfrac{12}{13}}{\dfrac{5}{13}} \]
\[ \Rightarrow \tan \left( a \right)=\dfrac{12}{5} \]
Also, $a={{\tan }^{-1}}\left( \dfrac{12}{5} \right)$
Now we will consider ${{\cos }^{-1}}\left( \dfrac{4}{5} \right)=b$. By taking the inverse cosine term to the right side of the equal sign we will have $\left( \dfrac{4}{5} \right)=\cos \left( b \right)$. Now as we know that $\cos \left( \theta \right)=\dfrac{\text{Base}}{\text{Hypotenuse}}$. Thus we have base as 4 units and hypotenuse as 5 units. By using Pythagoras theorem here we will find the value of the perpendicular. Therefore, we have the following diagram.
Thus by using Pythagoras theorem we have
\[ {{\left( 5 \right)}^{2}}={{y}^{2}}+{{\left( 4 \right)}^{2}} \]
\[ \Rightarrow {{y}^{2}}={{\left( 5 \right)}^{2}}-{{\left( 4 \right)}^{2}} \]
\[ \Rightarrow {{y}^{2}}=25-16 \]
\[ \Rightarrow {{y}^{2}}=9 \]
\[ \Rightarrow y=\pm 3 \]
As we know that the side cannot be negative. Therefore, we have that y = 3 which is the perpendicular in the triangle. Now will find the value of $\sin \left( b \right)=\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}$. Thus after substituting the values we have $\sin \left( b \right)=\dfrac{3}{5}$. As we know that $\tan \left( b \right)=\dfrac{\sin \left( b \right)}{\cos \left( b \right)}$. Therefore we have
\[ \tan \left( b \right)=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}} \]
\[ \Rightarrow \tan \left( b \right)=\dfrac{3}{4} \]
Also, $b={{\tan }^{-1}}\left( \dfrac{3}{4} \right)$.
Now, we will apply the formula given by $\tan \left( a+b \right)=\dfrac{\tan \left( a \right)+\tan \left( b \right)}{1-\tan \left( a \right)\tan \left( b \right)}$. After substituting the values we have
\[ \tan \left( a+b \right)=\dfrac{\dfrac{12}{5}+\dfrac{3}{4}}{1-\left( \dfrac{12}{5} \right)\dfrac{3}{4}} \]
\[ \Rightarrow \tan \left( a+b \right)=\dfrac{\dfrac{48+15}{20}}{\dfrac{20-36}{20}} \]
\[ \Rightarrow \tan \left( a+b \right)=\dfrac{\dfrac{63}{20}}{\dfrac{-16}{20}} \]
\[ \Rightarrow \tan \left( a+b \right)=-\dfrac{63}{16} \]
By taking the inverse term tan to the right side of the equation we can have $\left( a+b \right)={{\tan }^{-1}}\left( -\dfrac{63}{16} \right)$.
As we know that the value of a and b in terms of tan are $a={{\tan }^{-1}}\left( \dfrac{12}{5} \right)$ and $b={{\tan }^{-1}}\left( \dfrac{3}{4} \right)$. Thus we can have ${{\tan }^{-1}}\left( \dfrac{12}{5} \right)+{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( -\dfrac{63}{16} \right)$. Now we will use the formula ${{\tan }^{-1}}\left( -x \right)=\pi -{{\tan }^{-1}}\left( x \right)$. Thus we have
\[ {{\tan }^{-1}}\left( \dfrac{12}{5} \right)+{{\tan }^{-1}}\left( \dfrac{3}{4} \right)={{\tan }^{-1}}\left( -\dfrac{63}{16} \right) \]
\[ \Rightarrow {{\tan }^{-1}}\left( \dfrac{12}{5} \right)+{{\tan }^{-1}}\left( \dfrac{3}{4} \right)=\pi -{{\tan }^{-1}}\left( \dfrac{63}{16} \right) \]
\[ \Rightarrow {{\tan }^{-1}}\left( \dfrac{12}{5} \right)+{{\tan }^{-1}}\left( \dfrac{3}{4} \right)+{{\tan }^{-1}}\left( \dfrac{63}{16} \right)=\pi \]
Hence, the expression ${{\sin }^{-1}}\left( \dfrac{12}{13} \right)+{{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\tan }^{-1}}\left( \dfrac{63}{16} \right)=\pi $ is proved.
Note: We could have solved it by an alternate method. In this method we can solve it as
\[ {{\sin }^{-1}}\left( \dfrac{12}{13} \right)=a \]
\[ \Rightarrow \left( \dfrac{12}{13} \right)=\sin \left( a \right) \]
By using the trigonometry identity here which is given by ${{\cos }^{2}}\left( a \right)+{{\sin }^{2}}\left( a \right)=1$. This results in ${{\cos }^{2}}\left( a \right)=1-{{\sin }^{2}}\left( a \right)$. Therefore we have
\[ {{\cos }^{2}}\left( a \right)=1-{{\sin }^{2}}\left( a \right) \]
\[ \Rightarrow {{\cos }^{2}}\left( a \right)=1-{{\left( \dfrac{12}{13} \right)}^{2}} \]
\[ \Rightarrow {{\cos }^{2}}\left( a \right)=1-\dfrac{144}{169} \]
\[ \Rightarrow {{\cos }^{2}}\left( a \right)=\dfrac{169-144}{169} \]
\[ \Rightarrow {{\cos }^{2}}\left( a \right)=\dfrac{25}{169} \]
\[ \Rightarrow \cos \left( a \right)=\pm \dfrac{5}{13} \]
We will choose here the value of $\cos \left( a \right)=\dfrac{5}{13}$.
Similarly, we can solve the remaining by this formula and get the desired proof.
