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**Hint:**We have inverse cosine functions in both the RHS and the LHS of the given expression, so first we will try to transform them. We can do so by taking the term \[co{{s}^{-1}}\left( \dfrac{4}{5} \right)\] as x and the term \[co{{s}^{-1}}\left( \dfrac{12}{13} \right)\] as y. Then we can apply the formula\[\cos (x+y)=\cos x.\cos y-\sin x.\sin y\]to simplify and solve further."

**Complete step by step answer:**

Here each term in LHS, as well as RHS, is given as the inverse of the cosine function. First of all, remove the inverse by taking cosine in the terms having the inverse of cosine.

According to the question we have to prove \[{{\cos }^{-1}}\dfrac{4}{5}\]+\[co{{s}^{-1}}\left( \dfrac{12}{13} \right)\]=\[co{{s}^{-1}}\dfrac{33}{65}\]

Now, we have to remove the inverse functions from the terms present in LHS.

Let us assume,

\[x={{\cos }^{-1}}\dfrac{4}{5}\] ………….. (1)

The RHS part in the question is given in inverse cosine function. So, we need to convert this inverse cosine into a cosine function.

Now, applying cosine in LHS as well as RHS in the equation, We get,

\[\begin{align}

& x={{\cos }^{-1}}\dfrac{4}{5} \\

& \Rightarrow \cos x=\dfrac{4}{5}. \\

\end{align}\]

Similarly, let us assume,

\[y={{\cos }^{-1}}\dfrac{12}{13}\] ………………(2)

Again applying cosine in LHS as well as RHS in the equation (2), we get

\[\begin{align}

& y={{\cos }^{-1}}\dfrac{12}{13} \\

& \Rightarrow \cos y=\dfrac{12}{13} \\

\end{align}\]

Putting the equation (1) and equation (2) in the equation, we get

\[\begin{align}

& {{\cos }^{-1}}\dfrac{4}{5}+{{\cos }^{-1}}\dfrac{12}{13} \\

& =x+y \\

\end{align}\]

This means we have to find the value of (x+y).

As the RHS in the question has the inverse of cosine, here, we have to represent the whole LHS part into one cosine function.

We also know that,

\[\cos (x+y)=\cos x.\cos y-\sin x.\sin y\].

We know the identity, \[{{\sin }^{2}}x+{{\cos }^{2}}x=1\] , using this identity we can get

\[\begin{align}

& \cos x=\dfrac{4}{5} \\

&\Rightarrow \sin x=\sqrt{1-{{\cos }^{2}}x} \\

& \\

\end{align}\]

Putting the value of “ \[\cos x\] ” we get,

\[\begin{align}

&\Rightarrow \sin x=\sqrt{1-\dfrac{16}{25}} \\

&\Rightarrow \sin x=\sqrt{\dfrac{9}{25}} \\

&\Rightarrow \sin x=\dfrac{3}{5}. \\

\end{align}\]

Similarly, using the value of “ \[\cos y\] ”, we can get the value of “\[siny\]”.

\[\begin{align}

&\Rightarrow \cos y=\dfrac{12}{13} \\

&\Rightarrow \sin y=\sqrt{1-{{\cos }^{2}}y} \\

\end{align}\]

Putting the value of “ \[\cos y\] ” we get,

\[\begin{align}

&\Rightarrow \sin \,y=\sqrt{1-\dfrac{144}{169}} \\

&\Rightarrow \sin x=\sqrt{\dfrac{25}{169}} \\

&\Rightarrow \sin x=\dfrac{5}{13}. \\

\end{align}\]

Applying the formula, we get \[\cos (x+y)\].

\[\begin{align}

&\Rightarrow \cos (x+y)=cosx.cosy-sinx.siny \\

&\Rightarrow cos(x+y)=\dfrac{4}{5}.\dfrac{12}{13}-\dfrac{3}{5}.\dfrac{5}{13} \\

&\Rightarrow \cos (x+y)=\dfrac{48}{65}-\dfrac{15}{65} \\

&\Rightarrow \cos (x+y)=\dfrac{33}{65} \\

& \Rightarrow x+y=co{{s}^{-1}}\dfrac{33}{65}. \\

\end{align}\]

Therefore, LHS=RHS.

Hence, proved.

**Note:**We can also solve this question by converting the inverse of cosine present in LHS as well as RHS into the inverse of tan. We can do so by taking \[{{\cos }^{-1}}\left( \dfrac{4}{5} \right)=x\] and \[co{{s}^{-1}}\left( \dfrac{12}{13} \right)=y\].

\[\begin{align}

& {{\cos }^{-1}}\left( \dfrac{4}{5} \right)=x \\

& \Rightarrow \cos x=\dfrac{4}{5} \\

\end{align}\]

Using Pythagoras theorem, we can find the height.

Height = \[\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( base \right)}^{2}}}\]

\[\begin{align}

&\Rightarrow Height =\sqrt{{{5}^{2}}-{{4}^{2}}} \\

&\Rightarrow Height =\sqrt{25-16} \\

&\Rightarrow Height =\sqrt{9} \\

&\Rightarrow Height=3 \\

\end{align}\]

\[\tan x=\dfrac{height}{base}\]

\[\Rightarrow\tan x=\dfrac{3}{4}\]

Similarly,

\[\begin{align}

& {{\cos }^{-1}}\left( \dfrac{12}{13} \right)=y \\

& \Rightarrow \cos y=\dfrac{12}{13} \\

\end{align}\]

Using Pythagoras theorem, we can find the height.

Height = \[\sqrt{{{\left( hypotenuse \right)}^{2}}-{{\left( base \right)}^{2}}}\]

\[\begin{align}

&\Rightarrow Height =\sqrt{{{13}^{2}}-{{12}^{2}}} \\

&\Rightarrow Height =\sqrt{169-144} \\

& \Rightarrow Height =\sqrt{25} \\

&\Rightarrow Height =5 \\

\end{align}\]

\[\tan x=\dfrac{height}{base}\]

\[\tan x=\dfrac{5}{12}\]

And then using the formula \[{{\tan }^{-1}}A+{{\tan }^{-1}}B={{\tan }^{-1}}\left( \dfrac{A+B}{1-AB} \right)\] , we can get the required result.

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