Question

solve this question

Answer 1

HELLO FRIEND HERE IS YOUR ANSWER,,,,,,,,,,

The following data has been provided in the question,

$= > x = r \sin(A) \cos(C) \\ \\ \\ = > y = r \sin(A) \sin(C) \\ \\ \\ = > z = r \cos(A)$

Here we are given a equation to prove both the sides for satisfying the conditions. We square both the sides to obtain a product of square in left hand side for easier understanding and equation solving.

Therefore, we have,,,

$= > r = \sqrt{ {x}^{2} + {y}^{2} + {z}^{2} } \\ \\ \\ = > {(r)}^{2} = { ( \sqrt{ {x}^{2} + {y}^{2} + {z}^{2} } ) }^{2} \\ \\ \\ = > {r}^{2} = {x}^{2} + {y}^{2} + {z}^{2}$

Now, by saving the right hand side to equal the left hand side to prove the following equation.

Substituting the respective variable values, we get,,,,

$= > {(r \sin(A) \cos(C)) }^{2} + {(r \sin(A) \sin(C) )}^{2} + {(r \cos(A)) }^{2}$

Now solving the whole right hand side by applying trigonometric identities....

$= r^2 \sin^2{A}\cos^2{C} + r^2 \sin^2{A}\sin^2{C} + r^2 \cos^2{A}$

Taking the required common out of the trigonometric function of sin from the bracket.

$= > r^2 \sin^2{A} ( \cos^2{C}+\sin^2{C} )+ r^2 \cos^2{A}$

By applying common trigonometric identity rule of; $\sin^2{C}+\cos^2{C} = 1$, we get,,

$r^2 \sin^2{A} + r^2 \cos^2{A}$

Taking the common variable out, that is $"r^2"$ out of the brackets , we get,,

$r^2(\sin^2{A} + \cos^2{A})$

By applying common trigonometric identity rule of; $\sin^2{C}+\cos^2{C} = 1$, we get,,

$r^2$

HENCE PROVEN THE BOTH SIDES TO BE EQUAL.

HOPE IT HELPS YOU AND CLEARS YOUR DOUBTS FOR APPLYING TRIGONOMETRIC IDENTITIES AND PROVING IT SUBSEQUENTLY!!!!!!