A) \[-\frac{1}{2}\]
B) \[\frac{1}{2}\]
C) ? 1
D) 1
Correct Answer: C
Solution :
\[f(x)=\frac{2{{\cos }^{2}}\frac{x}{2}-2\sin \frac{x}{2}\cos \frac{x}{2}}{2{{\cos }^{2}}\frac{x}{2}+2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}\] \[=\tan \left( \frac{\pi }{4}-\frac{x}{2} \right)\] at \[x=\pi \], \[f(\pi )=-\tan \frac{\pi }{4}=-1\].
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