Question

Prove the following ${{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right)$ .

Answer 1

Hint: In the given question we are asked to prove L.H.S = R.H.S. To prove that first we need to find the sum of ${{\cos }^{-1}}\left( \dfrac{4}{5} \right)$ and ${{\cos }^{-1}}\left( \dfrac{12}{13} \right)$ with the help of formula i.e. ${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right)$ to get the answer.

Complete step-by-step answer:
To prove that ${{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right)$ let us take $x=\dfrac{4}{5}$ and $y=\dfrac{12}{13}$ .
We know that ${{\cos }^{-1}}x+{{\cos }^{-1}}y={{\cos }^{-1}}\left( xy-\sqrt{1-{{x}^{2}}}\sqrt{1-{{y}^{2}}} \right)$.
We have,
${{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)$.
$={{\cos }^{-1}}\left( \dfrac{4}{5}\times \dfrac{12}{13}-\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}\sqrt{1-{{\left( \dfrac{12}{13} \right)}^{2}}} \right)$.
$={{\cos }^{-1}}\left( \dfrac{48}{65}-\sqrt{1-\dfrac{16}{25}}\sqrt{1-\dfrac{144}{169}} \right)$ .
$={{\cos }^{-1}}\left( \dfrac{48}{65}-\sqrt{\dfrac{25-16}{25}}\sqrt{\dfrac{169-144}{169}} \right)$.
$={{\cos }^{-1}}\left( \dfrac{48}{65}-\left( \dfrac{3}{5}\times \dfrac{5}{13} \right) \right)$ .
$={{\cos }^{-1}}\left( \dfrac{48}{65}-\dfrac{15}{65} \right)$ .
$={{\cos }^{-1}}\left( \dfrac{48-15}{65} \right)$ .
$={{\cos }^{-1}}\left( \dfrac{33}{65} \right)$ .
Hence proved,
${{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right)$ .

Note: There’s an alternative way to solve this question,
Let $a={{\cos }^{-1}}\dfrac{4}{5}$ and \[b={{\cos }^{-1}}\dfrac{12}{13}\]
Let $a={{\cos }^{-1}}\dfrac{4}{5}$
$\cos a=\dfrac{4}{5}$
We know that ,
$\begin{align}
  & {{\sin }^{2}}a=1-{{\cos }^{2}}a \\
 & \sin a=\sqrt{1-{{\cos }^{2}}a} \\
 & =\sqrt{1-{{\left( \dfrac{4}{5} \right)}^{2}}}=\dfrac{3}{5} \\
\end{align}$
Let $b={{\cos }^{-1}}\dfrac{12}{13}$
$\cos b=\dfrac{12}{13}$
We know that ,
$\begin{align}
  & {{\sin }^{2}}b=1-{{\cos }^{2}}b \\
 & \sin b=\sqrt{1-{{\cos }^{2}}b} \\
 & =\sqrt{1-{{\left( \dfrac{12}{13} \right)}^{2}}}=\dfrac{5}{13} \\
\end{align}$
We know that $\cos \left( a+b \right)=\cos a\cos -\sin a\sin b$
Put the values $\cos a=\dfrac{4}{5},\cos b=\dfrac{12}{13},\sin a=\dfrac{3}{5}\text{and }\sin b=\dfrac{5}{13}$ .
$\begin{align}
  & \cos \left( a+b \right)=\dfrac{4}{5}\times \dfrac{12}{13}-\dfrac{3}{5}\times \dfrac{5}{13} \\
 & =\dfrac{48-15}{65}=\dfrac{33}{65} \\
 & \therefore \cos \left( a+b \right)=\dfrac{33}{65} \\
\end{align}$
$a+b={{\cos }^{-1}}\left( \dfrac{33}{65} \right)$
${{\cos }^{-1}}\left( \dfrac{4}{5} \right)+{{\cos }^{-1}}\left( \dfrac{12}{13} \right)={{\cos }^{-1}}\left( \dfrac{33}{65} \right)$
Hence proved.