Question

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 :

• A. xy - z = (x + y)z
• B. xy + z = (x + y)z
• C. xy + yz + zx = z
• D. xyz = 4

Hint:

Remember the basic identity $\sin^2 \theta + \cos^2 \theta = 1$ to relate terms in trigonometric series.

Answer 1

The Correct Option is B

Step 1: Sum of infinite G.P. $S = \frac{1}{1-r}$. $x = \frac{1}{1-\cos^2 \theta} = \frac{1}{\sin^2 \theta} \Rightarrow \sin^2 \theta = 1/x$. $y = \frac{1}{1-\sin^2 \phi} = \frac{1}{\cos^2 \phi} \Rightarrow \cos^2 \phi = 1/y$.
Step 2: $z = \frac{1}{1 - \cos^2 \theta \sin^2 \phi} = \frac{1}{1 - (1-1/x)(1-1/y)}$.
Step 3: $z = \frac{1}{1 - (1 - 1/x - 1/y + 1/xy)} = \frac{1}{1/x + 1/y - 1/xy} = \frac{xy}{x+y-1}$.
Step 4: $z(x+y-1) = xy \Rightarrow xz + yz - z = xy \Rightarrow xy + z = (x+y)z$.