Question

Given that $0<x<\frac{\pi }{4}$ and $\frac{\pi }{4}<y<\frac{\pi }{2}$ and ${\sum }_{k=0}^{\infty }(-1{)}^k{\tan }^{2k}x=p;{\sum }_{k=1}^{\infty }(-1{)}^k{\cot }^{2k}y=q;$ then ${\sum }_{k=1}^{\infty }{\tan }^{2k}x{\cot }^{2k}y$ is

• A. $\frac{1}{p}+\frac{1}{q}-\frac{1}{pq}$
• B. $\frac{1}{\{\frac{1}{p}+\frac{1}{q}-\frac{1}{pq}\}}$
• C. $p+q-pq$
• D. $p+q+pq$

Answer 1

The correct option is B $\frac{1}{\{\frac{1}{p}+\frac{1}{q}-\frac{1}{pq}\}}$

$p=$ Infinite G.P.
Where $a=1,r=-{\tan }^{2}x$
$\therefore p=\frac{a}{1-r}=\frac{1}{1+ta{n}^{2}x}={\cos }^{2}x,q=\frac{1}{1+{\cot }^{2}y}=si{n}^{2}y$
$\therefore S=\frac{1}{1-{\tan }^{2}x{\cot }^{2}y}=\frac{1}{1-\left(\frac{1-{\cos }^{2}x}{{\cos }^{2}x}\right)\left(\frac{1-si{n}^{2}y}{si{n}^{2}y}\right)}$
$S=\frac{pq}{p+q-1}=\frac{1}{\{\frac{1}{p}+\frac{1}{q}-\frac{1}{pq}\}}$