Question

If x = r sinA cosC , y = r sinA sinC , z = r cosA ;

then prove that r² = x² + y² + z²​

Answer 1

Step-by-step explanation:

Given:

• x = r sinA cosC
• y = r sinA sinC
• z = r cosA

To Prove:

• r² = x² + y² + z²

Solution: Taking RHS , we have

• x² + y² + z²

[ Now put the value of x , y and z in RHS ]

➮ x² = (r sinA cosC)²

• x² = r² sin²A cos²C

➮ y² = (r sinA sinC)²

• y² = r² sin²A sin²C

➮ z² = (r cosA)²

• z² = r² cos²A

By adding above terms

➟ r² sin²A cos²C + r² sin²A sin²C + r² cos²A

➟ r² sin²A ( cos²C + sin²C ) + r² cos² A

➟ r² sin²A ( 1 ) + r² cos² A

➟ r² sin²A + r² cos²A

➟ r² ( sin²A + cos²A )

➟ r² × 1

➟ r²

Hence, LHS = RHS i.e r² = x² + y² + z².

★ Identities used here

• sin²θ + cos²θ = 1

$\large\bold{\texttt {Proved }}$

Answer 2

Answer:

Given :-

⬤ x = r sinA cosC

⬤ y = r sinA sinC

⬤ z = r cosA

To Prove :-

⬤ r² = x² + y² +z²

Solution :-

➣ R.H.S = x² + y² + z²

✫ Putting the value of x, y and z in R.H.S we get,

$\implies$ (r sinA cosC)² + (r sinA sinC)² + (r cosA)²

$\implies$ r² sin²A cos²C + r² sin²A sin²C + r² cos ²A

$\implies$ r² sin²A(cos²C + sin²C) + r² cos ²A

$\implies$ r² sin²A(1) + r² cos²A

$\implies$ r² sin²A + r² cos²A

$\implies$ r²(sin²A + cos²A)

$\implies$ r²(1)

➨ r² = L.H.S

$\mapsto$ $\boxed{\bold{\large{PROVED}}}$

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