Question

x = 4cos theeta - 5 sin theeta , y = 4sin theeta + 5cos théeta eliminate the theeta​

Answer 1

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

Given that,

$\rm \: x = 4cos\theta - 5sin\theta - - - (1)$

and

$\rm \: y = 4sin\theta + 5cos\theta - - - (2)$

Now, Consider

$\rm \: {x}^{2}$

$\rm \: = \: {(4cos\theta - 5sin\theta )}^{2}$

$\rm \: = \: {16cos}^{2}\theta + 25 {sin}^{2}\theta - 2 \times 4cos\theta \times 5sin\theta$

$\rm \: = \: {16cos}^{2}\theta + 25 {sin}^{2}\theta - 40sin\theta cos\theta$

So,

$\rm\implies \: {x}^{2} = \: {16cos}^{2}\theta + 25 {sin}^{2}\theta - 40sin\theta cos\theta - - (3)$

Now, Consider

$\rm \: {y}^{2}$

$\rm \: = \: {(4sin\theta + 5cos\theta )}^{2}$

$\rm \: = \: {16sin}^{2}\theta + 25 {cos}^{2}\theta + 2 \times 4sin\theta \times 5cos\theta$

$\rm \: = \: {16sin}^{2}\theta + 25 {cos}^{2}\theta + 40sin\theta cos\theta$

So,

$\rm\implies \: {y}^{2} = \: {16sin}^{2}\theta + 25 {cos}^{2}\theta + 40sin\theta cos\theta - - - (4)$

On adding equation (3) and (4), we get

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

$\rm \: = \: 16( {sin}^{2}\theta + {cos}^{2}\theta ) + 25( {sin}^{2}\theta + {cos}^{2}\theta )$

$\rm \: = \: 16 \times 1 + 25 \times 1$

$\rm \: = \: 16 + 25$

$\rm \: = \: 41$

$\rm\implies \: {x}^{2} + {y}^{2} = 41$

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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

Answer 2

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

We have,

• x = 4cosθ - 5sinθ —— (1)
• y = 4sinθ + 5cosθ —— (2)

Opening brackets by squaring,

x² = (4cosθ - 5sinθ)²

x² = 16cos²θ + 25sin²θ - 2 × 4cosθ × 5sinθ

x² = 16cos²θ + 25sin²θ - 40sinθcosθ

Opening brackets by squaring,

y² = (4sinθ + 5cosθ)²

y² = 16sin²θ + 25cos²θ + 2 × 4sinθ × 5cosθ

y² = 16sin²θ + 25cos²θ + 40sinθcosθ

Adding on both the sides,

x² + y² = 16 (sin²θ + cos²θ) + 25 (sin²θ + cos²θ)

x² + y² = 16 × 1 + 25 × 1 —— (sin²θ + cos²θ = 1)

x² + y² = 16 + 25

x² + y² = 41