Question

If x = a cos , y = b sin , then b2 x 2 + a2 y 2 – a 2b 2 is equal to
​

Answer 1

x = acos∅

y = bsin∅

→ b²x² + a²y² - a²b²

= b²a²cos²∅ + a²b²sin²∅ - a²b²

= a²b²(cos²∅ + sin²∅) - a²b²

= a²b² - a²b²

= 0

Identity used :

cos²∅ + sin²∅ = 1

Answer 2

Question :

If x = a cos θ, y = b sin θ, then b² x² + a² y² – a²b²

How to do :

First according to the given condition we need to put the values and then we will take

Solution :

• b² x² + a² y² – a²b²

Putting the values we get :

• b² (a cos θ)² + a² (b sin θ)² - a²b²

• b² a² cos² θ + a² b² sin² θ - a²b²

By taking a²b² as common we get

• a²b²(cos² θ + sin² θ) - a²b²

Now according to the square relations we know that sin² θ + cos² θ = 1

• a²b²(1) - a²b²

• a²b² - a²b²

• 0

Hence, the answer is 0

Know More :

Square Relations :

• sin² θ + cos² θ = 1

• sec² θ – tan² θ = 1

• cosec² θ – cot² θ= 1

Quotient Relations :

• sin θ× cosec θ = 1

• cos θ × sec θ = 1

• tan θ × cot θ = 1

Basic :

• sin ∅ = P/H

• cos ∅ = B/H

• tan ∅ = P/B

• cot = B/P

• sec = H/B

• cosec = H/P

Here,

P refers Perpendicular or Height

B refers Base

H refers Hypotentuse

Trigonometric value of standard angles :

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered} \end{gathered}$

Trigonometrical ratios of complementary angles :

• sin (90° - θ) = cosec θ

• cos (90° - θ) = sec θ

• tan (90° - θ) = cot θ

• cot (90° - θ) = tan θ

• sec (90° - θ) = cos θ

• cosec (90° - θ) = sin θ

Regards

# BeBrainly