sqrt 2 + sqrt 3 + sqrt 4 + sqrt 6 = . . .. | vedclass

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Question

(a) We have $\cot A = \frac{{\cos A}}{{\sin A}} $

$= \frac{{2{{\cos }^2}A}}{{2\sin A\cos A}} = \frac{{1 + \cos 2A}}{{\sin 2A}}$

Putting $A = 7\f

Vedclass · Accepted Answer

$\cot 7\frac{{{1^o}}}{2}$

Vedclass · Answer

(a) We have $\cot A = \frac{{\cos A}}{{\sin A}} $

$= \frac{{2{{\cos }^2}A}}{{2\sin A\cos A}} = \frac{{1 + \cos 2A}}{{\sin 2A}}$

Putting $A = 7\frac{{{1^o}}}{2} $

$\Rightarrow \cot 7\frac{{{1^o}}}{2} = \frac{{1 + \cos {{15}^o}}}{{\sin {{15}^o}}}$

On simplification, we get

$\cot 7\frac{{{1^o}}}{2} = \sqrt 6 + \sqrt 2 + \sqrt 3 + \sqrt 4 $.

$\sqrt 2 + \sqrt 3 + \sqrt 4 + \sqrt 6 = . . ..$

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