Question

if x=r sinA cos B,y= r sin A*and ,z=r cos A..prove x²+y²+z²=r²​

Answer 1

Appropriate Question :-

If x = r sinA cosB, y = r sinA sinB and z = r cosA, prove that

x²+y²+z²=r²

$\large\underline{\sf{Solution-}}$

Given that,

x = r sinA cosB

y = r sinA sinB

z = r cosA

Now,

Consider, LHS

$\rm :\longmapsto\: {x}^{2} + {y}^{2} + {z}^{2}$

On substituting the values of x, y and z we get

$\rm \: = {(rsinAcosB)}^{2} + {(rsinAsinB)}^{2} + {(rcosA)}^{2}$

$\rm \: = \: \: {r}^{2} {sin}^{2}A {cos}^{2} B + {r}^{2} {sin}^{2}A {sin}^{2}B + {r}^{2} {cos}^{2}A$

$\rm \: = \: \: {r}^{2} {sin}^{2}A( {cos}^{2} B + {sin}^{2}B )+ {r}^{2} {cos}^{2}A$

We know,

$\boxed{ \bf{ \: {sin}^{2}x + {cos}^{2}x = 1}}$

So, using this, the above reduces to

$\rm \: = \: \: {r}^{2} {sin}^{2}A(1 )+ {r}^{2} {cos}^{2}A$

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

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

$\rm \: = \: \: {r}^{2} (1)$

$\rm \: = \: \: {r}^{2}$

Hence,

$\boxed{ \bf{ \: \: {x}^{2} + {y}^{2} + {z}^{2} = {r}^{2} }}$

Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1