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Prove that $\cot \dfrac{\pi }{4} = \sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $

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Hint: Here in this question we must know the following properties and identities of trigonometric functions.
They are mentioned below: -
$\cot A = \dfrac{{(1 + \cos 2A)}}{{\sin 2A}}$
Conversion of radian into degree is done by multiplying that angle by $\dfrac{{180}}{\pi }$
$\cos (A - B) = \cos A\cos B + \sin A\sin B$

Complete step-by-step answer:
First of all we will convert the angle which is given in radian into degree so that the calculation part can become a little less confusing.
$\cot \dfrac{\pi }{4} = \cot (\dfrac{\pi }{4} \times \dfrac{{180}}{\pi })$
$ \Rightarrow \cot \dfrac{{180}}{4} = \cot 7.5 = \cot \dfrac{{15}}{2}$
Now we will apply $\cot A = \dfrac{{(1 + \cos 2A)}}{{\sin 2A}}$ identity so that fractional part can be eliminated. $\cot \dfrac{{15}}{2} = \dfrac{{(1 + \cos 2 \times \dfrac{{15}}{2})}}{{\sin 2 \times \dfrac{{15}}{2}}}$
$ \Rightarrow \dfrac{{(1 + \cos 15)}}{{\sin 15}}$ (Cancelling 2 from numerator and denominator)
For finding the value of $\cos 15$ we know that $\cos 15 = \cos (45 - 30)$ so applying identity $\cos (A - B) = \cos A\cos B + \sin A\sin B$
 $ \Rightarrow \cos (45 - 30) = \cos 45\cos 30 + \sin 45\sin 30$ (Here A=45 and B=30)
$ \Rightarrow \dfrac{1}{{\sqrt 2 }} \times \dfrac{{\sqrt 3 }}{2} + \dfrac{1}{{\sqrt 2 }} \times \dfrac{1}{2}$ (Putting values of trigonometric functions)
$ \Rightarrow \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$ (Taking L.C.M)
$ \Rightarrow \dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }}$ (Rationalising by multiplying and dividing $\sqrt 2 $)
$ \Rightarrow \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{{2 \times 2}} = \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4}$
$\therefore \cos 15 = \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4}$
Now putting value of $\cos 15$in $\dfrac{{(1 + \cos 15)}}{{\sin 15}}$
$ \Rightarrow \dfrac{{(1 + \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\sin 15}}$
For $\sin 15$ we will use $\cos 15$ because $\cos \theta = \dfrac{b}{h}$and \[\sin \theta = \dfrac{p}{h}\]where p=perpendicular, b=base and h=hypotenuse.
$\therefore \cos 15 = \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4} = \dfrac{b}{h}$0
\[h = \sqrt {{p^2} + {b^2}} \](By Pythagoras theorem)
\[ \Rightarrow p = \sqrt {{h^2} - {b^2}} \]
\[ \Rightarrow p = \sqrt {{4^2} - {{[\sqrt 2 (\sqrt 3 + 1])}^2}} \]
Further simplifying we will get
\[ \Rightarrow p = \sqrt {16 - [2{{(3 + 1 + 2\sqrt 3 ]}^2})} \] (Expanding\[{(a + b)^2} = {a^2} + {b^2} + 2ab\])
\[ \Rightarrow p = \sqrt {16 - [2(4 + 2\sqrt 3 ])} \]
\[ \Rightarrow p = \sqrt {16 - [8 + 4\sqrt 3 ]} \] (Multiplying 2 inside)
\[ \Rightarrow p = \sqrt {16 - 8 - 4\sqrt 3 } \]
\[ \Rightarrow p = \sqrt {8 - 4\sqrt 3 } \] (Taking 4 outside the root)
\[\therefore p = 2\sqrt {2 - \sqrt 3 } \]
\[ \Rightarrow p = \dfrac{{2\sqrt {2 - \sqrt 3 } }}{4}\]Or $\dfrac{{\sqrt 2 (\sqrt 3 - 1)}}{4}$ because square root of \[\sqrt {2 - \sqrt 3 } \] is $\dfrac{{(\sqrt 3 - 1)}}{{\sqrt 2 }}$
You can directly write sine value from cosine value without doing so much of calculation by just making a sign opposite.
Now putting values of $\sin 15$ in $\dfrac{{(1 + \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\sin 15}}$ we will get
$ \Rightarrow \dfrac{{(1 + \dfrac{{\sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\dfrac{{\sqrt 2 (\sqrt 3 - 1)}}{4}}}$
$ \Rightarrow \dfrac{{(\dfrac{{4 + \sqrt 2 (\sqrt 3 + 1)}}{4})}}{{\dfrac{{\sqrt 2 (\sqrt 3 - 1)}}{4}}}$ (Cancelling 4 from both denominator terms)\[ \Rightarrow \dfrac{{4 + \sqrt 2 (\sqrt 3 + 1)}}{{\sqrt 2 (\sqrt 3 - 1)}} = \dfrac{{4 + \sqrt 6 + \sqrt 2 }}{{\sqrt 6 - \sqrt 2 }}\]
Now rationalising the denominator irrational term
\[ \Rightarrow \dfrac{{4 + \sqrt 6 + \sqrt 2 }}{{\sqrt 6 - \sqrt 2 }} \times \dfrac{{\sqrt 6 + \sqrt 2 }}{{\sqrt 6 + \sqrt 2 }}\]\[ \Rightarrow \dfrac{{(4 + \sqrt 6 + \sqrt 2 )(\sqrt 6 + \sqrt 2 )}}{{6 - 2}} = \dfrac{{4\sqrt 6 + 4\sqrt 2 + 6 + \sqrt 6 \times \sqrt 2 + \sqrt 6 \times \sqrt 2 + 2}}{4}\]\[ \Rightarrow \dfrac{{4\sqrt 6 + 4\sqrt 2 + 8 + 2(\sqrt {12} )}}{4} = \dfrac{{4\sqrt 6 + 4\sqrt 2 + 8 + 2(2\sqrt 3 )}}{4}\]\[ \Rightarrow \dfrac{{4\sqrt 6 + 4\sqrt 2 + 8 + 4\sqrt 3 }}{4} = \sqrt 6 + \sqrt 2 + 2 + \sqrt 3 \]
\[\therefore \sqrt 6 + \sqrt 2 + \sqrt 4 + \sqrt 3 \] (2 can be written as $\sqrt 4 $)
Therefore $\cot \dfrac{\pi }{4} = \sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 $
Hence proved.

Note: Students should know basic values of trigonometric functions which can boost their calculation part some of the values are as follows: -
$\cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2},\sin {30^ \circ } = \dfrac{1}{2}$
$\cos {60^ \circ } = \dfrac{1}{2},\sin {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
$\cos {45^ \circ } = \dfrac{1}{{\sqrt 2 }},\sin {45^ \circ } = \dfrac{1}{{\sqrt 2 }}$
\[\cos {90^ \circ } = 0,\sin {90^ \circ } = 1\]
$\cos {180^ \circ } = 1,\sin {180^ \circ } = 0$
Also while doing rationalisation some students make mistakes by multiplying with the same term but that is not correct. The same term with opposite signs should be multiplied and divided.