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