Question

x=rsinAcosA y=rsinAsinC z=rcosA prove that r²=x²+y²z²(without spam)​

Answer 1

Step-by-step explanation:

Given,

x = rsinAcosC ------1

y = rsinAsinC -----2

z = rcosA -------3

squaring eq 1, 2 and 3, we get

x² = r²sin²Acos²C ----4

y² = r²sin²Asin²C -----5

z² = r²cos²A ------6

Adding 4, 5 and 6, we get,

x²+ y²+ z²= r²sin²Acos²C+ r²sin²Asin²C+ r²cos²A

x²+ y²+ z²= r²sin²A(cos²C +sin²C)+ r²cos²A

x²+ y²+ z²= r²sin²A+ r²cos²A -----[sin²∅+cos²∅=1]

x²+ y²+ z²= r²(sin²A+ cos²A)

x²+ y²+ z²= r² ------- [sin²∅+cos²∅=1]

Hence, proved.

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