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

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Sequences and Series

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

and

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

then :

Answer: (D)

Each sum is geometric with ratio in

(0, 1)

, so

x = \frac{1}{1 - cos ^{2} θ} = \frac{1}{sin ^{2} θ}, y = \frac{1}{1 - sin ^{2} ϕ} = \frac{1}{cos ^{2} ϕ}, z = \frac{1}{1 - cos ^{2} θ sin ^{2} ϕ} .

Put

p = cos^{2} θ

and

q = sin^{2} ϕ

, so

x = \frac{1}{1 - p}

y = \frac{1}{1 - q}

and

z = \frac{1}{1 - pq}

. Then

x y + z = \frac{( 1 - pq ) + ( 1 - p ) ( 1 - q )}{( 1 - p ) ( 1 - q ) ( 1 - pq )} = \frac{2 - p - q}{( 1 - p ) ( 1 - q ) ( 1 - pq )},

and since

x + y = \frac{2 - p - q}{( 1 - p ) ( 1 - q )}

, the product

(x + y) z

has exactly this value. Hence

x y + z = (x + y) z

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Single correct option