If x=sum_{n=0}^{infty} cos ^{2 n} theta,y=sum | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $x = \sum_{n=0}^{\infty} \cos^{2n} 	heta = \frac{1}{1 - \cos^2 	heta} = \frac{1}{\sin^2 	heta} \implies \sin^2 	heta = \frac{1}{x}$.Give

Vedclass · Accepted Answer

$x z+y z=x y+z$

Vedclass · Answer

(A) Given $x = \sum_{n=0}^{\infty} \cos^{2n} 	heta = \frac{1}{1 - \cos^2 	heta} = \frac{1}{\sin^2 	heta} \implies \sin^2 	heta = \frac{1}{x}$.Given $y = \sum_{n=0}^{\infty} \sin^{2n} 	heta = \frac{1}{1 - \sin^2 	heta} = \frac{1}{\cos^2 	heta} \implies \cos^2 	heta = \frac{1}{y}$.Since $\sin^2 	heta + \cos^2 	heta = 1$,we have $\frac{1}{x} + \frac{1}{y} = 1$,which implies $x + y = xy$.Given $z = \sum_{n=0}^{\infty} (\cos^2 	heta \sin^2 	heta)^n = \frac{1}{1 - \cos^2 	heta \sin^2 	heta} = \frac{1}{1 - \frac{1}{x} \cdot \frac{1}{y}} = \frac{xy}{xy - 1}$.From $xy - 1 = \frac{xy}{z}$,we have $xy - 1 = \frac{x+y}{z}$ (since $xy = x+y$).Thus,$z(xy - 1) = x + y \implies xyz - z = x + y$.Since $x+y = xy$,we can write $xyz - z = xy$,or $xyz = xy + z$.Alternatively,using $xy = x+y$,we have $z(x+y) = xy + z$,which simplifies to $xz + yz = xy + z$.

If $x=\sum_{n=0}^{\infty} \cos ^{2 n} \theta$,$y=\sum_{n=0}^{\infty} \sin ^{2 n} \theta$,$z=\sum_{n=0}^{\infty} \cos ^{2 n} \theta \sin ^{2 n} \theta$ and $0 < \theta < \frac{\pi}{2}$,then

Similar Questions

If all the terms of an $A.P.$ are squared,then the new series will be in

If $A_1, A_2$ are two arithmetic means,$G_1, G_2$ are two geometric means,and $H_1, H_2$ are two harmonic means between two numbers,then $\frac{A_1 + A_2}{H_1 + H_2} \cdot \frac{H_1 H_2}{G_1 G_2} = \dots$

Let $S$ be the sum of the first $9$ terms of the series: $(x+ka) + (x^2+(k+2)a) + (x^3+(k+4)a) + (x^4+(k+6)a) + \ldots$ where $a \neq 0$ and $x \neq 1$. If $S = \frac{x^{10}-x+45a(x-1)}{x-1}$,then $k$ is equal to:

If three unequal numbers $p, q, r$ are in $H.P.$ and their squares are in $A.P.$,then the ratio $p:q:r$ is

If $a, b, c$ are in $G.P.$ and $x, y$ are the arithmetic means between $a, b$ and $b, c$ respectively,then $\frac{a}{x} + \frac{c}{y}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module