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**Hint:**We know that \[y = \sin x = \cos x\] when \[x = \dfrac{\pi }{4}\] use this to find the slopes of both the curves then usen the formula \[\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\] where $m_1$ and $m_2$ are slopes of given curves.

**Complete step-by-step answer:**Let the two curves be \[{C_1}\] and \[{C_2}\] respectively

\[{C_1}:y = \sin x,{C _ 2}:y = \cos x\]

Equation both the curves we are getting

\[\begin{array}{l}

\sin x = \cos x\\

\therefore x = \dfrac{\pi }{4}

\end{array}\]

Therefore curves intersect each other at the point P : \[x = \dfrac{\pi }{4}\]

Now, differentiating with respect to. x,

\[{C_1}\] gives : \[\dfrac{{dy}}{{dx}} = \cos x\]

\[{C_2}\] gives : \[\dfrac{{dy}}{{dx}} = - \sin x\]

Hence the slopes \[{{m_1}\& {m_2}}\] of \[{C_1}\& {C_2}\] at P: \[x = \dfrac{\pi }{4}\] are

\[\begin{array}{l}

{m_1} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\\

{m_2} = - \sin \dfrac{\pi }{4} = - \dfrac{1}{{\sqrt 2 }}

\end{array}\]

If \[\theta \] is the acute angle between them at P, then

\[\begin{array}{l}

\therefore \tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\\

\Rightarrow \tan \theta = \left| {\dfrac{{\left( {\dfrac{1}{{\sqrt 2 }}} \right) - \left( { - \dfrac{1}{{\sqrt 2 }}} \right)}}{{1 + \left( {\dfrac{1}{{\sqrt 2 }}} \right)\left( { - \dfrac{1}{{\sqrt 2 }}} \right)}}} \right|\\

\Rightarrow \tan \theta = \dfrac{{\left| {2\left( {\dfrac{1}{{\sqrt 2 }}} \right)} \right|}}{{\left| {\dfrac{{\left( {2 - 1} \right)}}{2}} \right|}}\\

\Rightarrow \tan \theta = 2\sqrt 2 \\

\therefore \theta = {\tan ^{ - 1}}\left( {2\sqrt 2 } \right)

\end{array}\]

**So it can be clearly seen that option a is the correct option here.**

**Note:**One must know the formula \[\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|\] in order to solve these types of questions also remember that if we differentiate any function we get the slope of the curve formed by that function.

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