For 0- - - -2- if x- -n-0-cos2n- y-n-0-sin2n- z-n-0-cos2n - then

X=1+cos^2 ϕ +cos^4 ϕ+....=1/(1 cos^2 ϕ)=1/sin^2 ϕ y=1+sin^2 ϕ+sin^4 ϕ+...=1/(1 sin^2 ϕ)=1/cos^2 ϕ z=1+cos^2 ϕ sin^2 ϕ+cos^4 ϕ sin^4 ϕ+..=1/(1 cos^2 ϕ sin^2 ϕ) ...

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Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

For 0< ϕ < π/...

Question

For 0< $ϕ$ < $\frac{π}{2}, i f x = \sum_{n = 0}^{\infty} c o s^{2 n} ϕ, y = \sum_{n = 0}^{\infty} s i n^{2 n} ϕ, z = \sum_{n = 0}^{\infty} c o s^{2 n} ϕ, t h e n$

xyz = xz + y

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xyz = xy + z

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xyz = x + y + z

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xyz = yz + x

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Solution

The correct option is C
xyz = x + y + z

$x = 1 + c o s^{2} ϕ + c o s^{4} ϕ + . . . . = \frac{1}{(1 - c o s^{2} ϕ)} = \frac{1}{s i n^{2} ϕ} y = 1 + s i n^{2} ϕ + s i n^{4} ϕ + . . . = \frac{1}{(1 - s i n^{2} ϕ)} = \frac{1}{c o s^{2} ϕ} z = 1 + c o s^{2} ϕ s i n^{2} ϕ + c o s^{4} ϕ s i n^{4} ϕ + . . = \frac{1}{(1 - c o s^{2} ϕ s i n^{2} ϕ)}$

$N o w x y z = \frac{1}{s i n^{2} ϕ c o s^{2} ϕ (1 - c o s^{2} ϕ s i n^{2} ϕ)} x y + z = \frac{1}{s i n^{2} ϕ c o s^{2} ϕ} + \frac{1}{1 - c o s^{2} ϕ s i n^{2} ϕ} = \frac{1}{s i n^{2} ϕ c o s^{2} ϕ (1 - c o s^{2} ϕ s i n^{2} ϕ)} = x y z$

which is given in (b)

Also x + y + z = xyz , which is given in (c).

Suggest Corrections

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For 0< ϕ < π2, if x=∑∞n=0cos2nϕ, y=∑∞n=0sin2nϕ, z=∑∞n=0cos2n ϕ, then

For 0< $ϕ$ < $\frac{π}{2}, i f x = \sum_{n = 0}^{\infty} c o s^{2 n} ϕ, y = \sum_{n = 0}^{\infty} s i n^{2 n} ϕ, z = \sum_{n = 0}^{\infty} c o s^{2 n} ϕ, t h e n$