Question

42.
If x sina cosB, y = sina sinB, z = cosa, x^2 + y^2+ z^2 is equal to
(A) 0
(B) 1
(C) A'B+ c
(D) 2

Answer 1

Appropriate Question

If x = sina cosb, y = sina sinb, z = cosa, then x² + y² + z² is

equal to

(A) 0

(B) 1

(C) ab + c

(D) 2

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

Given that

$\rm :\longmapsto\: x \: = \: sina \: cosb$

$\rm :\longmapsto\: y \: = \: sina \: sinb$

$\rm :\longmapsto\:z \: = \: cosa$

Now Consider

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

$\rm \: = \: {(sina \: cosb)}^{2} + {(sina \: sinb)}^{2} + {(cosa)}^{2}$

$\rm \: = \: {sin}^{2}a \: {cos}^{2}b + {sin}^{2}a \: {sin}^{2}b \: + \: {cos}^{2} a$

$\rm \: = \: ( {sin}^{2}a \: {cos}^{2}b + {sin}^{2}a \: {sin}^{2}b )\: + \: {cos}^{2} a$

$\rm \: = \: {sin}^{2}a \: ( {cos}^{2}b + \: {sin}^{2}b )\: + \: {cos}^{2} a$

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

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

$\rm \: = \: 1$

Hence,

$\rm :\longmapsto\: \boxed{ \tt{ \: {x}^{2} + {y}^{2} + {z}^{2} = 1 \: }}$

• So, option (B) is correct

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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