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**Hint:**Here in this question some of the trigonometric identities will get used which are being mentioned below: -

$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$

$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$

$\tan {45^ \circ } = 1$

**Complete step-by-step answer:**

$\dfrac{{\cos {{11}^ \circ } + \sin {{11}^ \circ }}}{{\cos {{11}^ \circ } - \sin {{11}^ \circ }}} = \tan {56^ \circ }$

L.H.S=$\dfrac{{\cos {{11}^ \circ } + \sin {{11}^ \circ }}}{{\cos {{11}^ \circ } - \sin {{11}^ \circ }}}$Now we will prove L.H.S (left hand side) equal to R.H.S (right hand side)

First of all we will divide numerator and denominator with $\cos {11^ \circ }$

$ \Rightarrow \dfrac{{\dfrac{{\cos {{11}^ \circ }}}{{\cos {{11}^ \circ }}} + \dfrac{{\sin {{11}^ \circ }}}{{\cos {{11}^ \circ }}}}}{{\dfrac{{\cos {{11}^ \circ }}}{{\cos {{11}^ \circ }}} - \dfrac{{\sin {{11}^ \circ }}}{{\cos {{11}^ \circ }}}}}$

Now we will apply identity $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$

$ \Rightarrow \dfrac{{\dfrac{{\cos {{11}^ \circ }}}{{\cos {{11}^ \circ }}} + \tan {{11}^ \circ }}}{{\dfrac{{\cos {{11}^ \circ }}}{{\cos {{11}^ \circ }}} - \tan {{11}^ \circ }}}$

Now cancelling $\cos {11^ \circ }$ from numerator and denominator

$ \Rightarrow \dfrac{{1 + \tan {{11}^ \circ }}}{{1 - \tan {{11}^ \circ }}}$

Now we will convert this equation such that we can apply identity $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$$ \Rightarrow \dfrac{{\tan {{45}^ \circ } + \tan {{11}^ \circ }}}{{1 - \tan {{45}^ \circ }\tan {{11}^ \circ }}}$ (As$\tan {45^ \circ } = 1$, Here A=45 and B =11)

Now we can apply identity $\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$

$ \Rightarrow \tan ({45^ \circ } + {11^ \circ })$ (Adding the inner angles)

$ \Rightarrow \tan ({56^ \circ })$

Therefore L.H.S=R.H.S i.e. $\dfrac{{\cos {{11}^ \circ } + \sin {{11}^ \circ }}}{{\cos {{11}^ \circ } - \sin {{11}^ \circ }}} = \tan {56^ \circ }$

**Additional Information:**Trigonometric functions are real functions which relate an angle of a right-angled triangle to ratios of two side length. Most widely used trigonometric functions are sine, cosine and tangent. The angles of sine, cosine and tangent are the primary classification of functions of trigonometry. And the functions which are cotangent, secant and cosecant can be derived from the primary functions. Behaviour of all these functions in four quadrants is as follows: -

First quadrant = All trigonometric functions are positive (sine, cosine, tan, sec, cosec, cot)

Second quadrant=Positive (sine, cosec), Negative (cosine, tan, sec, cot)

Third quadrant= Positive (tan, cot), Negative (sine, cosine, sec, cosec)

Fourth quadrant= Positive (cosine, sec), Negative (sine, tan, cot, cosec)

**Note:**Students may likely to make one common mistake in this question is that they will try to think the direct value of given trigonometric functions which is a very wrong way to approach because one cannot memorise too many values rather some specific values should be remembered to solve these types of questions. Some of the values are mentioned below: -

$\tan {30^ \circ } = \dfrac{1}{{\sqrt 3 }}$

$\tan {60^ \circ } = \sqrt 3 $

$\tan {45^ \circ } = 1$

\[\tan {90^ \circ }\]=infinity

$\tan {180^ \circ } = 0$

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