A) \[2007\]
B) \[2006\]
C) \[2009\]
D) \[2008\]
Correct Answer: D
Solution :
\[{{\theta }_{1}}+{{\theta }_{2}}=\frac{\pi }{2}\] \[\therefore \] \[I=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{\frac{d\theta }{1+\tan \left( \frac{\pi }{2}-\theta \right)}=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{\frac{\tan \theta \,\,d\theta }{1+\tan \theta }}}\] and also\[I=\int_{{{\theta }_{1}}}^{{{\theta }_{2}}}{\frac{d\theta }{1+\tan \theta }}\] \[\therefore \]\[2I=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{d\theta ={{\theta }_{2}}-{{\theta }_{1}}}=\frac{1002\pi }{2008}\Rightarrow I=\frac{501\pi }{2008}\] Hence\[,\]\[K=2008\].
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