A) \[\frac{1}{p}+\frac{1}{q}-\frac{1}{pq}\]
B) \[\frac{1}{\left\{ \frac{1}{p}+\frac{1}{q}-\frac{1}{pq} \right\}}\]
C) \[p+q-pq\]
D) \[p+q+pq\]
Correct Answer: B
Solution :
[b]: p = Infinite G.P. where \[a=1,r=-{{\tan }^{2}}x\] \[\therefore \]\[p=\frac{a}{1-r}=\frac{1}{1+{{\tan }^{2}}x}={{\cos }^{2}}x\] Similarly, \[q=\frac{1}{1+{{\cot }^{2}}y}={{\sin }^{2}}y\] Now, \[\sum\limits_{k=0}^{\infty }{{{\tan }^{2k}}x{{\cot }^{2k}}}y=\frac{1}{1-{{\tan }^{2}}x{{\cot }^{2}}y}\] \[=\frac{1}{1-\left( \frac{1-{{\cos }^{2}}x}{{{\cos }^{2}}x} \right)\left( \frac{1-{{\sin }^{2}}y}{{{\sin }^{2}}y} \right)}\] \[=\frac{pq}{p+q-1}=\frac{1}{\left\{ \frac{1}{p}+\frac{1}{q}-\frac{1}{pq} \right\}}\]You need to login to perform this action.
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