A) \[\sqrt{6}+\sqrt{3}+\sqrt{2}-2\]
B) \[\sqrt{6}-\sqrt{3}+\sqrt{2}-2\]
C) \[\sqrt{6}-\sqrt{3}+\sqrt{2}+2\]
D) \[\sqrt{6}-\sqrt{3}-\sqrt{2}-2\]
Correct Answer: B
Solution :
We have \[\tan A=\frac{\sin A}{\cos A}=\frac{2\sin A\cos A}{2{{\cos }^{2}}A}=\frac{\sin 2A}{1+{{\cos }^{2}}A}\] Putting \[A=7{{\frac{1}{2}}^{o}}\Rightarrow \tan 7{{\frac{1}{2}}^{o}}=\frac{\sin {{15}^{o}}}{1+\cos {{15}^{o}}}\] On simplification, we get\[\tan 7{{\frac{1}{2}}^{o}}=\sqrt{6}-\sqrt{3}+\sqrt{2}-2\].
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