If \[\mathbf{x}=\mathbf{6}{{\mathbf{0}}^{{}^\circ }}\]then \[\frac{1}{2}\sqrt{1+cosx}+\frac{1}{2}\sqrt{1-cosx}\] is equal to

If \[\mathbf{x}=\mathbf{6}{{\mathbf{0}}^{{}^\circ }}\]then \[\frac{1}{

1.	If \[\mathbf{x}=\mathbf{6}{{\mathbf{0}}^{{}^\circ }}\]then \[\frac{1}{2}\sqrt{1+cosx}+\frac{1}{2}\sqrt{1-cosx}\] is equal to
A.	\[cot\frac{\theta }{2}\]
B.	\[\sec \frac{\theta }{2}\]
C.	\[\sin \frac{\theta }{2}\]
D.	\[\cos \frac{\theta }{2}\]
Answer» E.

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If \[\mathbf{x}=\mathbf{6}{{\mathbf{0}}^{{}^\circ }}\]then \[\frac{1}{2}\sqrt{1+cosx}+\frac{1}{2}\sqrt{1-cosx}\] is equal to

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