• # question_answer $\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{6}$ is equal to [IIT 1966, 1975] A) $\cot 7\frac{{{1}^{o}}}{2}$ B) $\sin 7\frac{{{1}^{o}}}{2}$ C) $\sin \,{{15}^{o}}$ D) $\cos \,\,{{15}^{o}}$

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 $\cot 7\frac{{{1}^{o}}}{2}=\sqrt{6}+\sqrt{2}+\sqrt{3}+\sqrt{4}$.