Question

# Statement: If $x = r\sin A\cos C,y = r\sin A\sin C$ and $z = r\cos A$, the ${r^2} = {x^2} + {y^2} + {z^2}$.State whether the given statement is A. TrueB. False

Hint: In this question, first of all write the given data and substitute these values in the given equation. Then take the common terms and use the trigonometry formula to prove the statement is true. So, use this concept to reach the solution of the given problem.

Given that $x = r\sin A\cos C,y = r\sin A\sin C$ and $z = r\cos A$
Here we have to prove ${r^2} = {x^2} + {y^2} + {z^2}$.
Now consider the value ${x^2} + {y^2} + {z^2}$
$\Rightarrow {x^2} + {y^2} + {z^2} = {\left( {r\sin A\cos C} \right)^2} + {\left( {r\sin A\sin C} \right)^2} + {\left( {r\cos A} \right)^2} \\ \Rightarrow {x^2} + {y^2} + {z^2} = {r^2}{\sin ^2}A{\cos ^2}C + {r^2}{\sin ^2}A{\sin ^2}C + {r^2}{\cos ^2}A \\$
Taking the terms common and using the formula ${\sin ^2}x + {\cos ^2}x = 1$, we get
$\Rightarrow {x^2} + {y^2} + {z^2} = {r^2}{\sin ^2}A\left( {{{\cos }^2}C + {{\sin }^2}C} \right) + {r^2}{\cos ^2}A \\ \Rightarrow {x^2} + {y^2} + {z^2} = {r^2}{\sin ^2}A\left( 1 \right) + {r^2}{\cos ^2}A{\text{ }}\left[ {\because {{\sin }^2}x + {{\cos }^2}x = 1} \right] \\ \Rightarrow {x^2} + {y^2} + {z^2} = {r^2}{\sin ^2}A + {r^2}{\cos ^2}A \\$
Again, taking the terms common and using the formula ${\sin ^2}x + {\cos ^2}x = 1$, we get
$\Rightarrow {x^2} + {y^2} + {z^2} = {r^2}\left( {{{\sin }^2}A + {{\cos }^2}A} \right) \\ \Rightarrow {x^2} + {y^2} + {z^2} = {r^2}{\text{ }}\left[ {\because {{\sin }^2}A + {{\cos }^2}A = 1} \right] \\ \therefore {x^2} + {y^2} + {z^2} = {r^2} \\$
Hence proved that ${x^2} + {y^2} + {z^2} = {r^2}$
Note: Here we have used the formula ${\sin ^2}x + {\cos ^2}x = 1$. To solve these kinds of problems, substitute the given values in the equation and find whether it satisfies or not. If it satisfies then the given statement is true otherwise false.