Question

if xsin³A+ycos³A=sinA.cosA and xsinA=ycosA prove that x²+y²=1​

Answer 1

Step-by-step explanation:

Given:-

• xsin³A + ycos³A = sinA.cosA

• x sinA = y cosA

To Prove:-

• x² + y² = 1

Proof:-

$\sf \implies x {sin}^{3} A = y {cos}^{3} A = sinA.cosA$

$\sf \implies (xsinA) \: {sin}^{2} A + (y cosA) \: {cos}^{2} A = sinA.cosA$

Since, x sinA = y cosA....(i)

$\sf \implies (xsinA) \: {sin}^{2} A + (xsinA) \: {cos}^{2} A = sinA.cosA$

$\sf \implies (xsinA) \big[{sin}^{2} A + {cos}^{2} A \big] = sinA.cosA$

$\sf \implies xsinA= sinA.cosA$

$\implies \boxed{\textsf{\textbf{x = cosA}}}$

Substitute x = cosA in (i)

$\sf \implies x sinA = y cosA$

$\sf \implies cosA.sinA = y cosA$

$\implies \boxed{\textsf{\textbf{y = sinA}}}$

Now, x² + y²

= sin²A + cos²A = 1

$\large \dashrightarrow \underline{\boxed{\therefore \bf{ \: {x}^{2} + {y}^{2} = 1}}}$

Hence, Proved

Answer 2

${\large{\underline{\boxed{\bf{\pink{Given:}}}}}}$

• xsin³A+ycos³A=sinA.cosA
• xsinA=ycosA

${\large{\underline{\boxed{\bf{\pink{To \: \: prove :}}}}}}$

• x²+y²=1

${\Large{\underline{\boxed{\bf{\red{Solution:}}}}}}$

⟹ xsin³A + ycos³A = sinAcosA.

⟹ (xsinA) sin²A + (ycosA) cos²A = sinAcosA.

We know that [ ycosA = xsinA].

⟹ (xsinA) sin²A + (xsinA) cos²A = sinAcosA

⟹ (xsinA) (sin²A + cos²A) = sinAcosA.

⟹ xsinA = sinAcosA(1).

Sin²A + Cos²A = 1

⟹ x = cosA

Squaring both side

⟹ x² = cos²A (2).

Now,

⟹ xsinA = ycosA.

⟹ cosAsinA = ycosA[ From equal(1)].

⟹ y = sinA

Squaring both side

⟹ y² = sin²A (3).

Adding equation (2) and (3).

⟹ x² + y² = sin²A + cos²A.

${\large{\underline{\boxed{\bf{\green{x² + y² = 1.}}}}}}$

Hence, it is proved