if x=a{1-(1/sinØ)} and y=a{(1/cosØ)+tanØ} prove that xy^2+a^2(2a-x)=0

Question

get me an answer plz

Answer 1

Question: if x=a{1-(1/sinØ)} and y=a{(1/cosØ)+tanØ} prove that xy^2+a^2(2a-x)=0.

2a - x = 2a - a(1 - 1/sinØ) = a(1 + 1/sinØ)

a^2(2a - x) = a^3(1 + 1/sinØ)

xy^2 = a(1 - 1/sinØ).a^2(1/cosØ + tanØ)^2

xy^2 = a^3.(1 - 1/sinØ).(1 + sinØ)^2/cos^2Ø

xy^2 = a^3.(1 - 1/sinØ).(1 + sinØ)^2/(1 - sin^2Ø)

xy^2 = a^3.([sinØ - 1]/sinØ).(1 + sinØ)^2/{(1 - sinØ)(1 + sinØ)}

xy^2 = a^3.([sinØ - 1]/sinØ).(1 + sinØ)/{(1 - sinØ)}

xy^2 = a^3.([-1]/sinØ).(1 + sinØ)

xy^2 = -a^3.(1 + sinØ)/sinØ

xy^2 = -a^3.(1/sinØ + 1)

xy^2 + a^2(2a - x) = -a^3.(1/sinØ + 1) + a^3(1 + 1/sinØ)

xy^2 + a^2(2a - x) = a^3.{-1/sinØ - 1 + 1 + 1/sinØ) = 0

xy^2 + a^2(2a - x) = 0

Answer 2

From the given equation, sinø ≠ 0, cosø ≠ 0 

x = a{1 - (1/sinø)} = a(sinø - 1) / sinø ··· Eq.1

y = a{(1/cosø) + tanø} = a{(1/cosø) + (sinø/cosø)} = a(1 + sinø) / cosø, so

y² = a²(1 + sinø)² / cos²ø = a²(1 + sinø)² / (1 - sinø²) = a²(1 + sinø)² / (1+ sinø)·(1 - sinø), thus

y² = a²(1 + sinø) / (1 - sinø) ··· Eq.2

Let sinø = X.   Plug X into Eq.1 and Eq.2, we have: x = a(X - 1) / X, y² = a²(1 + X) / (1 - X), so

xy² = {a(X - 1) / X}·{a²(1 + X) / (1 - X)} = -a³(1 + X) / X ··· Eq.3 

a²(2a - x) = a³{2 - (X - 1) / X} = a³(X + 1) / X··· Eq.4   

From Eq.3 and Eq.4, we have: xy² + a²(2a - x) = -a³{(1 + X) / X} + a³{(X + 1) / X} = 0

Therefore, xy² + a²(2a - x) = 0