If x - -n - 0 -1ntan2n- and y - -n - 0cos2n- where 0 - - - -4- then-

Y = 1 + cos^2 θ + cos^4 θ +⋯ nbsp; ⇒ y = 11 cos^2θ⇒1y = sin^2 θ x = 1 tan^2θ + tan^4θ ⋯ nbsp; ⇒ x = 11 ( tan^2 θ) = cos^2θ ∴ x + 1y = 1 ⇒ y(1 x) = ...

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Byju's Answer

Standard XIII

Mathematics

Sum of Infinite Terms of a GP

If x = ∑∞n = ...

Question

If $x = \infty \sum n = 0 (- 1)^{n} {tan}^{2 n} θ$ and $y = \infty \sum n = 0 {cos}^{2 n} θ,$ where $0 < θ < \frac{π}{4},$ then:

y (1 + x) = 1

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x (1 - y) = 1

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y (1 - x) = 1

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x (1 + y) = 1

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Solution

The correct option is C $y (1 - x) = 1$
$y = 1 + {cos}^{2} θ + {cos}^{4} θ + \dots$
$\Rightarrow y = \frac{1}{1 - {cos}^{2} θ} \Rightarrow \frac{1}{y} = {sin}^{2} θ$
$x = 1 - {tan}^{2} θ + {tan}^{4} θ - \dots$
$\Rightarrow x = \frac{1}{1 - (- {tan}^{2} θ)} = {cos}^{2} θ$
$∴ x + \frac{1}{y} = 1 \Rightarrow y (1 - x) = 1$

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Similar questions

Q. If

x = \infty \sum n = 0 (- 1)^{n} {tan}^{2 n} θ

and

y = \infty \sum n = 0 {cos}^{2 n} θ,

where

0 < θ < \frac{π}{4},

then:

Q. If

0 < θ, ϕ < \frac{π}{2},

x = \infty \sum n = 0 {cos}^{2 n} θ,

y = \infty \sum n = 0 {sin}^{2 n} ϕ

and

z = \infty \sum n = 0 {cos}^{2 n} θ \cdot {sin}^{2 n} ϕ

then :

Q. If

0 < θ < \frac{π}{2}

and

x = \sum_{n = 0}^{\infty} {cos}^{2 n} θ, y = \sum_{n = 0}^{\infty} {sin}^{2 n} θ, z = \sum_{n = 0}^{\infty} {cos}^{2 n} θ {sin}^{2 n} θ

, then the value of

x y z

Q. If

x = \infty \sum n = 0 a^{n}, y = \infty \sum n = 0 b^{n}, z = \infty \sum n = 0 {(a b)}^{n}

,
where

a, b < 1

then prove that

x z + y z = x y + z

.
Another form:
For

0 < θ, ϕ - \frac{π}{2}

x = \infty \sum n = 0 {cos}^{2 n} θ

y = \infty \sum n = 0 {sin}^{2 n} ϕ, z = \infty \sum n = 0 {cos}^{2 n} θ {sin}^{2 n} ϕ

then prove that

x z + y z - z = x y

Q. For

0 < θ < \frac{π}{4}, x = \infty Σ n = 0 {sin}^{2 n} θ, y = \infty Σ n = 0 {cos}^{2 n} θ

, then match the entries of column -I with column - II

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If x=∞∑n=0(−1)ntan2nθ and y=∞∑n=0cos2nθ, where 0<θ<π4, then:

The correct option is C y(1−x)=1y=1+cos2θ+cos4θ+⋯ ⇒ y=11−cos2θ⇒1y=sin2θ x=1−tan2θ+tan4θ−⋯ ⇒ x=11−(−tan2θ)=cos2θ ∴ x+1y=1⇒y(1−x)=1

If $x = \infty \sum n = 0 (- 1)^{n} {tan}^{2 n} θ$ and $y = \infty \sum n = 0 {cos}^{2 n} θ,$ where $0 < θ < \frac{π}{4},$ then: