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Arun sharma

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· 1 Months ago

CONFUSED (Question:117122.0)

WHY DIVIDE BY SINX SINY SINZ

Question:
If sin2x + sin2y + sin2 z = (sinx + siny + sinz)2 then which one of the following expressions must necessarily follow? [\(x\neq0, y\neq0 \ and\ z\neq0\)]
Options:
A) tan x + tany + tan z= 0
B) \(\cfrac{1}{sin\ x}+\cfrac{1}{sin\ y}+\cfrac{1}{sin\ z} = 0\)
C) cos x + cos y +cosz = 0
D) \(\cfrac{1}{cos\ x}+\cfrac{1}{cos\ y}+\cfrac{1}{cos\ z} = 0\)
Solution:
Ans: (b) sin2x + sin2y + sin2z = (sinx + siny + sinz)2

= sin2x + sin2y + sin2z + 2 sinx siny + 2siny sinz + 2sinz sinx

\(\Rightarrow\)sinx siny + siny sinz + sinz sinx = 0

On Dividing by sinx siny sinz,

\(\cfrac{1}{sin\ z}+\cfrac{1}{sin\ x}+\cfrac{1}{sin\ y}=0\)

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