Question

For $0<\theta <\frac{\pi }{4},x=\stackrel{\infty }{\underset{n=0}{\Sigma }}{\sin }^{2n}\theta ,y=\stackrel{\infty }{\underset{n=0}{\Sigma }}{\cos }^{2n}\theta$, then match the entries of column -I with column - II

Answer 1

$x=$sum of an infinite G.P. whose sum by

$S_{\infty }=\frac{a}{1-r}$ is

$x=\frac{1}{1-{\sin }^2\theta }=\frac{1}{{\cos }^2\theta },y=\frac{1}{1-{\cos }^2\theta }=\frac{1}{{\sin }^2\theta }$

$\therefore {\cos }^2\theta =\frac{1}{x},{\sin }^2\theta =\frac{1}{y}$

(a) $\frac{1}{1-{\sin }^2\theta {\cos }^2\theta }=\frac{1}{1-\frac{1}{xy}}=\frac{xy}{xy-1}\to \left(r\right)$

(b) $\frac{1}{1-{\tan }^2\theta }=\frac{1}{1-\frac{1}{xy}}=\frac{y}{y-x}\to \left(q\right)$

(c) $\frac{1}{1-{\sin }^2\theta {\cos }^4\theta }=\frac{1}{1-\frac{1}{y}.\frac{1}{x^2}}=\frac{x^{2}y}{x^{2}y-1}\to \left(s\right)$

(d) $\frac{1}{1-{\sin }^2\theta {\cos }^4\theta }=\frac{1}{1-\frac{1}{x}.\frac{1}{y^2}}=\frac{x{y}^2}{x{y}^2-1}\to \left(p\right)$