Let 0- x-y-z -2 be such that sin x sin y cos z -12-2- sin2xsin3ycos z-14-2 and sin x sin4ycos2z-18- Then which of the following is-are CORRECT-

Given, sin x sin y cos z=12√(2) nbsp; …(1) sin^2x sin^3ycos z=14√(2) …(2) nbsp; nbsp; sin x sin^4ycos^2z=18 …(3) nbsp; nbsp; nbsp; ...

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Byju's Answer

Standard XII

Mathematics

AM,GM,HM Inequality

Let 0≤ x,y,z ...

Question

Let $0 \leq x, y, z \leq \frac{π}{2}$ be such that $sin x sin y cos z = \frac{1}{2 \sqrt{2}},$ $\frac{{sin}^{2} x {sin}^{3} y}{cos z} = \frac{1}{4 \sqrt{2}}$ and $\frac{sin x {sin}^{4} y}{{cos}^{2} z} = \frac{1}{8} .$ Then which of the following is/are CORRECT?

sin x + sin y = \frac{3}{2}

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cos (x - y) = \frac{\sqrt{3}}{2}

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tan (x + y + z) = \sqrt{3} - 2

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sec (y - z) = \sqrt{6} - \sqrt{2}

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Solution

The correct option is D $sec (y - z) = \sqrt{6} - \sqrt{2}$
Given, $sin x sin y cos z = \frac{1}{2 \sqrt{2}} \dots (1)$
$\frac{{sin}^{2} x {sin}^{3} y}{cos z} = \frac{1}{4 \sqrt{2}} \dots (2)$
$\frac{sin x {sin}^{4} y}{{cos}^{2} z} = \frac{1}{8} \dots (3)$

Now, $\frac{(1) \times (2)}{(3)}$
$\Rightarrow \frac{{sin}^{3} x {sin}^{4} y {cos}^{2} z}{sin x \cdot {sin}^{4} y} = \frac{1}{2} \Rightarrow {sin}^{2} x \cdot {cos}^{2} z = \frac{1}{2} \dots (4)$

Now, $\frac{(1)^{2}}{(4)}$
$\Rightarrow \frac{{sin}^{2} x {sin}^{2} y {cos}^{2} z}{{sin}^{2} x {cos}^{2} z} = \frac{1}{8} \cdot \frac{2}{1} \Rightarrow {sin}^{2} y = \frac{1}{4} \Rightarrow sin y = \frac{1}{2} [∵ y \in (0, \frac{π}{2})] \Rightarrow y = \frac{π}{6}$

From equation $(1),$
$sin x cos z = \frac{1}{\sqrt{2}} \dots (5)$
and from equation $(2),$
$\frac{{sin}^{2} x}{cos z} = \sqrt{2} \dots (6)$
From $(5) \times (6),$
$sin x = 1 \Rightarrow x = \frac{π}{2}$
From equation $(5),$
$z = \frac{π}{4}$

$∴ sin y = \frac{1}{2}, cos z = \frac{1}{\sqrt{2}}, sin x = 1$
and $x = \frac{π}{2}; y = \frac{π}{6}; z = \frac{π}{4}$

$sin x + sin y = 1 + \frac{1}{2} = \frac{3}{2}$

$cos (x - y) = cos \frac{π}{3} = \frac{1}{2}$

$x + y + z = \frac{11 π}{12} tan \frac{11 π}{12} = tan (π - \frac{π}{12}) = - tan \frac{π}{12} = \sqrt{3} - 2 sec (y - z) = sec \frac{π}{12}$

$cos (\frac{π}{4} - \frac{π}{6}) = cos \frac{π}{6} cos \frac{π}{4} + sin \frac{π}{6} sin \frac{π}{4} \Rightarrow cos (\frac{π}{12}) = \frac{\sqrt{3}}{2} \cdot \frac{1}{\sqrt{2}} + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \Rightarrow cos (\frac{π}{12}) = \frac{\sqrt{3} + 1}{2 \sqrt{2}} \Rightarrow sec (\frac{π}{12}) = \frac{2 \sqrt{2}}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \Rightarrow sec (\frac{π}{12}) = \sqrt{6} - \sqrt{2}$

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Similar questions

For x, y, z $ϵ (0, \frac{π}{2}),$ let x, y, z be first three consecutive terms of an arithmetic progression such that cos x + cox y + cos z = 1 and sin x + sin y + sin z = $\frac{1}{\sqrt{2}}$ , then which of the following is/are correct?

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Q. If

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Let a, b, x and y be real numbers such that $a - b = 1$ and $y \neq 0$ . If the complex numbers $z = x + i y$ satisfies Im $(\frac{a z + b}{z + 1}) = y$ , then which of the following is/are possible value(s) of x?

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Let 0≤x,y,z≤π2 be such that sinxsinycosz=12√2, sin2xsin3ycosz=14√2 and sinxsin4ycos2z=18. Then which of the following is/are CORRECT?

Let $0 \leq x, y, z \leq \frac{π}{2}$ be such that $sin x sin y cos z = \frac{1}{2 \sqrt{2}},$ $\frac{{sin}^{2} x {sin}^{3} y}{cos z} = \frac{1}{4 \sqrt{2}}$ and $\frac{sin x {sin}^{4} y}{{cos}^{2} z} = \frac{1}{8} .$ Then which of the following is/are CORRECT?