If x=sum_{n=0}^{infty}(-1)^{n} tan ^{2 n} the | vedclass

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

Question

(C) The given series for $x$ is a geometric series with first term $a=1$ and common ratio $r=-	an^2 	heta$.
Since $0 < 	heta < \frac{\pi}{4}

Vedclass · Accepted Answer

$y(1-x)=1$

Vedclass · Answer

(C) The given series for $x$ is a geometric series with first term $a=1$ and common ratio $r=-	an^2 	heta$.
Since $0 < 	heta < \frac{\pi}{4}$,we have $0 < 	an^2 	heta < 1$,so the series converges to $x = \frac{1}{1 - (-	an^2 	heta)} = \frac{1}{1 + 	an^2 	heta} = \frac{1}{\sec^2 	heta} = \cos^2 	heta$.
The given series for $y$ is a geometric series with first term $a=1$ and common ratio $r=\cos^2 	heta$.
Since $0 < 	heta < \frac{\pi}{4}$,we have $\frac{1}{2} < \cos^2 	heta < 1$,so the series converges to $y = \frac{1}{1 - \cos^2 	heta} = \frac{1}{\sin^2 	heta}$.
From the expression for $x$,we have $\cos^2 	heta = x$,which implies $\sin^2 	heta = 1 - x$.
Substituting this into the expression for $y$,we get $y = \frac{1}{1 - x}$.
Rearranging this gives $y(1 - x) = 1$.

If $x=\sum_{n=0}^{\infty}(-1)^{n} \tan ^{2 n} \theta$ and $y=\sum_{n=0}^{\infty} \cos ^{2 n} \theta,$ for $0 < \theta < \frac{\pi}{4},$ then

Similar Questions

Let $E = x^{2017} + y^{2017} + z^{2017} - 2017xyz$ (where $x, y, z \geq 0$),then the least value of $E$ is

The sum of three consecutive terms in a geometric progression is $14$. If $1$ is added to the first and the second terms and $1$ is subtracted from the third,the resulting new terms are in arithmetic progression. Then the lowest of the original terms is

If $f(x + y, x - y) = xy$,then the arithmetic mean of $f(x, y)$ and $f(y, x)$ is

If $x_1, x_2, \dots, x_n$ and $\frac{1}{h_1}, \frac{1}{h_2}, \dots, \frac{1}{h_n}$ are two $A.P.s$ such that $x_3 = h_2 = 8$ and $x_8 = h_7 = 20$,then $x_5 \cdot h_{10}$ equals

If the sum of two extreme numbers of an $A.P.$ with four terms is $8$ and the product of the remaining two middle terms is $15$, then the greatest number of the series will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module