A) 1
B) \[1+{{\sin }^{2}}1\]
C) \[1+{{\cos }^{2}}1\]
D) None of these
Correct Answer: B
Solution :
Let \[f(\theta )=co{{s}^{2}}\,(\cos \,\theta )+{{\sin }^{2}}(\sin \,\theta )\] \[\because \] \[-1\le \,\cos \,\theta \le 1\] and \[-1\le \sin \,\theta \le 1\] \[\therefore \] \[\cos \,1\le \,\cos \,(\cos \theta )\,\le 1\] and \[-\sin 1\le sin\,(sin\,\theta )\,\le \,sin\,1\] \[\therefore \] \[{{\cos }^{2}}1\le \,{{\cos }^{2}}\,(\cos \,\theta )\,\le 1\] and \[0\le {{\sin }^{2}}(\sin \,\theta )\,\le {{\sin }^{2}}1\] \[\therefore \] Maximum value \[=1+{{\sin }^{2}}1\]
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