If x = r sinA cosC, y = r sinA sinC and z = r cosA, then find the value of $$\text{x}^2+\text{y}^2+\text{z}^2$$​​

Correct option is A

Given:

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

then $$\text{x}^2+\text{y}^2+\text{z}^2$$

Formula Used:

$$sin^2\theta + cos^2 \theta =1$$

Solution:

​$$\text{x}^2+\text{y}^2+\text{z}^2$$= (r sinA cosC)²+ (r sinA sinC)²+  (r cosA)²

=r²sin²A cos²C + r^2​ sin²A sin²C + r²cos²A

=r²sin²A{cos²C +  sin²C}+ r²cos²A

=r²sin²A{1}+ r²cos²A

=r²{sin²A+ cos²A}

=r²{1}= r²

Option (A) is right.