Question

(xSinA-yCosA)^2+(xCosA+yCosA)^2=x^2+y^2

Answer 1

$\large\underline{\sf{To\:prove- }}$

$\sf \: {(xsinA - ycosA)}^{2} + {(xcosA + ysinA)}^{2} = {x}^{2} + {y}^{2}$

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

Consider LHS

$\rm :\longmapsto\:\sf \: {(xsinA - ycosA)}^{2} + {(xcosA + ysinA)}^{2}$

We know

$\boxed{ \tt{ \: \: {(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \: \: }}$

and

$\boxed{ \tt{ \: \: {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy \: \: }}$

So, on applying these Identities, we get

$\sf\: = {x}^{2} {sin}^{2}A + {y}^{2} {cos}^{2}A - 2xysinAcosA + {x}^{2} {cos}^{2}A + {y}^{2} {sin}^{2}A + 2xysinAcosA$

$\sf\: = {x}^{2} {sin}^{2}A + {y}^{2} {cos}^{2}A - \cancel{2xysinAcosA} + {x}^{2} {cos}^{2}A + {y}^{2} {sin}^{2}A + \cancel{ 2xysinAcosA}$

$\sf \:  =  \: {x}^{2} {sin}^{2}A + {y}^{2} {cos}^{2}A + {x}^{2} {cos}^{2}A + {y}^{2} {sin}^{2}A$

$\sf \:  =  \: {x}^{2} ({sin}^{2}A + {cos}^{2}A )+ {y}^{2}( {cos}^{2}A +{sin}^{2}A)$

We know,

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

So, using this identity, we get

$\rm \:  =  \: {x}^{2} \times 1 + {y}^{2} \times 1$

$\sf \:  =  \: {x}^{2} + {y}^{2}$

Hence,

$\boxed{ \tt{ \: {(xsinA - ycosA)}^{2} + {(xcosA + ysinA)}^{2} = {x}^{2} + {y}^{2} }}$

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