If 0- -2- x- n-0-cos2n- y- n-0-sin2n- and z- n-0-cos2n-sin2n- then xyz

Explanation for correct option:Step 1: Simplify x=∑ _n=0^∞cos^2n(θ).Given, x=∑ _n=0^∞cos^2n(θ)⇒ x=1+cos^2(θ)+cos^4(θ)+cos^6(θ)+.......(are in G.P)⇒ x=1/ ...

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Byju's Answer

Standard XII

Mathematics

Principal Solution of Trigonometric Equation

If 0<θ <π/2, ...

Question

If $0 < θ < \frac{π}{2}, x = \sum_{n = 0}^{\infty} \cos^{2 n} (θ), y = \sum_{n = 0}^{\infty} \sin^{2 n} (θ),$ and $z = \sum_{n = 0}^{\infty} \cos^{2 n} (θ) \sin^{2 n} (θ)$ , then $x y z$

$x y z = x z + y$

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$x y z = x y + z$

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$x y z = y z + x$

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none of these

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Solution

The correct option is B
$x y z = x y + z$

Explanation for correct option:
Step-1: Simplify $x = \sum_{n = 0}^{\infty} \cos^{2 n} (θ)$ .
Given, $x = \sum_{n = 0}^{\infty} \cos^{2 n} (θ)$
$\Rightarrow$ $x = 1 + \cos^{2} (θ) + \cos^{4} (θ) + \cos^{6} (θ) + . . . . . . .$ (are in G.P)
$\Rightarrow$ $x = \frac{1}{1 - \cos^{2} (θ)} (∵ S_{\infty} = \frac{a}{1 - r})$
$\Rightarrow$ $x = \frac{1}{\sin^{2} (θ)} . . . . . . . . . (i)$
Step-2: Simplify $y = \sum_{n = 0}^{\infty} \sin^{2 n} (θ)$ .
Given, $y = \sum_{n = 0}^{\infty} \sin^{2 n} (θ)$
$\Rightarrow$ $y = 1 + \sin^{2} (θ) + \sin^{4} (θ) + \sin^{6} (θ) + . . . . . . .$ (are in G.P)
$\Rightarrow$ $y = \frac{1}{1 - \sin^{2} (θ)} (∵ S_{\infty} = \frac{a}{1 - r})$
$\Rightarrow$ $y = \frac{1}{\cos^{2} (θ)} . . . . . . . . . (i i)$
Step-3: Simplify $z = \sum_{n = 0}^{\infty} \cos^{2 n} (θ) \sin^{2 n} (θ)$
$\Rightarrow$ $z = \sum_{n = 0}^{\infty} \cos^{2 n} (θ) \sin^{2 n} (θ)$
$\Rightarrow$ $z = 1 + \cos (θ) \sin (θ) + \cos^{2} (θ) \sin^{2} (θ) + \cos^{3} (θ) \sin^{3} (θ) + . . . . . . . . .$ (are in G.P)
$\Rightarrow$ $z = \frac{1}{1 - \sin^{2} (θ) \cos^{2} (θ)} . . . . . . . . . . (i i i)$
Step-4: Solving equation $(i), (i i) & (i i i)$ .
$\Rightarrow$ $z = \frac{1}{(1 - \frac{1}{x y})}$
$\Rightarrow$ $z = \frac{x y}{x y - 1}$
$\Rightarrow$ $x y = x y z - z$
$\Rightarrow$ $x y z = x y + z$
Hence correct answer is option $(B)$ .

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If 0<θ<π2,x=∑n=0∞cos2nθ,y=∑n=0∞sin2nθ, and z=∑n=0∞cos2nθsin2nθ, then xyz

If $0 < θ < \frac{π}{2}, x = \sum_{n = 0}^{\infty} \cos^{2 n} (θ), y = \sum_{n = 0}^{\infty} \sin^{2 n} (θ),$ and $z = \sum_{n = 0}^{\infty} \cos^{2 n} (θ) \sin^{2 n} (θ)$ , then $x y z$