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**Hint:**In order to solve this problem, we need to know multiple numbers of trigonometric identities and formula. The formulas we need to know are as follows, $\sin 2x=2\sin x\times \cos x$, $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$, ${{\sin }^{2}}1+{{\cos }^{2}}1=1$ , ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ and $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a.\tan b}$ and $\tan \left( -x \right)=-\tan x$.

**Complete step-by-step solution:**We are given the expression and we need to find the value of it.

The expression is ${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)$ .

In order to simplify this, we need to know certain identities.

The identities are $\sin 2x=2\sin x\times \cos x$

And $\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$

Using these identities, we get,

${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{2\sin 1.\cos 1-1}{{{\cos }^{2}}1-{{\sin }^{2}}1} \right)$

We can now use the rule that ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ in the denominator, we get,

${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{2\sin 1.\cos 1-1}{\left( \cos 1+\sin 1 \right)\left( \cos 1-\sin 1 \right)} \right)$

We can write 1 as ${{\sin }^{2}}1+{{\cos }^{2}}1$ because of the identity that ${{\sin }^{2}}1+{{\cos }^{2}}1=1$ .

The equation becomes,

${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{2\sin 1.\cos 1-\left( {{\sin }^{2}}1+{{\cos }^{2}}1 \right)}{\left( \cos 1+\sin 1 \right)\left( \cos 1-\sin 1 \right)} \right)$

In the numerator, we can combine all the terms using the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$ ,

We get,

${{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{-{{\left( \cos 1-\sin 1 \right)}^{2}}}{\left( \cos 1+\sin 1 \right)\left( \cos 1-\sin 1 \right)} \right)$

We can cancel few of the terms form numerator as well as the denominator.

After solving we get,

\[{{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{-\left( \cos 1-\sin 1 \right)}{\left( \cos 1+\sin 1 \right)} \right)\]

Taking the cos 1 common from the numerator we get,

\[\begin{align}

& {{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{-\cos 1\left( 1-\tan 1 \right)}{\cos 1\left( 1+\tan 1 \right)} \right) \\

& ={{\tan }^{-1}}\left( \dfrac{-\left( 1-\tan 1 \right)}{\left( 1+\tan 1 \right)} \right)

\end{align}\]

We knew that $\tan \dfrac{\pi }{4}=1$ , substituting we get,

\[{{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{-\left( \tan \dfrac{\pi }{4}-\tan 1 \right)}{\left( 1+\tan \dfrac{\pi }{4}.\tan 1 \right)} \right)\]

We can see that \[\dfrac{\left( \tan \dfrac{\pi }{4}-\tan 1 \right)}{\left( 1+\tan \dfrac{\pi }{4}.\tan 1 \right)}={{\tan }^{-1}}\left( \dfrac{\pi }{4}-1 \right)\].

We can use identity $\tan \left( a-b \right)=\dfrac{\tan a-\tan b}{1+\tan a.\tan b}$ .

Substituting we get,

\[{{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( -\tan \left( \dfrac{\pi }{4}-1 \right) \right)\]

We need to take the negative inside.

We must know the property $\tan \left( -x \right)=-\tan x$ .

Solving this we get,

\[\begin{align}

& {{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \tan \left( 1-\dfrac{\pi }{4} \right) \right) \\

& =1-\dfrac{\pi }{4}

\end{align}\]

**Hence, the correct option is (c).**

**Note:**In this problem, the main trick is to know which formula to apply when. Also, many tend to make the wrong approach. Another approach that is usually taken is as follows,\[\begin{align}

& {{\tan }^{-1}}\left( \dfrac{\sin 2-1}{\cos 2} \right)={{\tan }^{-1}}\left( \dfrac{\sin 2}{\cos 2}-\dfrac{1}{\cos 2} \right) \\

& ={{\tan }^{-1}}\left( \tan 2-\sec 2 \right) \\

& =2-{{\tan }^{-1}}\left( \sec 2 \right)

\end{align}\]

But this is not the correct approach to take.

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