Question

x = a(sino - cos) y = b(sino + cose) 30,
x2b2 + y2a2 = ?​

Answer 1

GIVEN:

• x=a(sinA-cosA)
• y=b(sinA+cosA)

$\LARGE{\bf{\underline{\underline{TO:FIND:-}}}}$

• x²b²+y²a²

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

Now,

$\to \sf x^2b^2+y^2a^2$

▣ Substitute the values.

$\sf [a(sinA-cosA)]^2 \cdot b^2+[b(sinA+cosA)]^2 \cdot a^2$

▣ Further simplify:

$\sf a^2(sin^2A+cos^2A-2sinAcosA) \cdot b^2+b^2(sin^2A+cos^2A+2sinAcosA) \cdot a^2$

$\to \sf a^2 b^2(sin^2A+cos^2A-2sinAcosA) +a^2b^2(sin^2A+cos^2A+2sinAcosA)$

▣ Take a²b² common.

$\to \sf a^2 b^2(sin^2A+cos^2A-2sinAcosA +sin^2A+cos^2A+2sinAcosA)$

▣ We know that sin²A+cos²A=1. Further simplifying:

$\to \sf a^2 b^2(1+1)$

$\leadsto \boxed{\sf{\red{2(a^2b^2)}}}$

∴x²b²+y²a²=2(a²b²)