If x sin θ – y cos θ = √(x2 + y2) and cos2 θ/a2 + sin2 θ/b2 = 1/(x2 + y2), then prove that y2/a2 + x2/b2 = 1

Trigonometric identities are equations that are true for every value of the variable in the domain and relate to various trigonometric functions. These are the qualities that exist for all values of the variables in the equation. Sine, cosine, tangent, cosecant, secant, and cotangent are the trigonometric ratios used in these identities. Right triangle sides, such as an adjacent side, opposite side, and hypotenuse side, are used to determine all of these trigonometric ratios. These trigonometric identities are only valid in a right-angle triangle. 

Sine Cosine identity

In trigonometry, the sine cosine identity is a Pythagorean trigonometric identity as it forms its basis on the Pythagoras theorem. It states that the sum of squares of sine and cosine for any angle θ is unity.

sin² θ + cos² θ = 1

Derivation

Consider a right triangle ABC with angle θ between its base and hypotenuse.

Applying Pythagoras theorem on this triangle, we get

AC² = AB² + BC²            

Dividing both sides by AC², we get

AC²/AC² = AB²/AC² + BC²/AC²

(AB/AC)² + (BC/AC)² = 1 ⇢ (1)

We know, for angle θ,

sin θ = Perpendicular/Hypotenuse

sin θ = AB/AC ⇢  (2)

Also, 

cos θ = Base/Hypotenuse

cos θ = BC/AC⇢ (3)

Using (2) and (3) in (1), 

sin² θ + cos² θ = 1

This proves the identity of sine and cosine ratios.

If x sin θ – y cos θ = √(x² + y²) and cos² θ/a² + sin² θ/b² = 1/(x² + y²), then prove that y²/a² + x²/b² = 1.

Solution:

x sin θ – y cos θ = √(x² + y²)

Squaring both sides we get,

=> x² sin² θ + y² cos² θ – 2xy sin θ cos θ = x² + y²

=> x² sin² θ + y² cos² θ – 2xy sin θ cos θ – x² – y² = 0

=> x² (sin² θ – 1) + y² (cos² θ -1) – 2xy sin θ cos θ = 0

=> – x² cos² θ – y² sin² θ – 2xy sin θ cos θ = 0

=> x² cos² θ + y² sin² θ + 2xy sin θ cos θ = 0

=> (x cos θ + y sin θ)² = 0

=> x cos θ = -y sin θ

=> x/sin θ = -y/cos θ

=> x²/sin² θ = y²/cos² θ

=> tan² θ = x²/y²

=> sin² θ = x²/(x² + y²) and cos² θ = y²/(x² + y²)

Put the above values in the equation cos² θ/a² + sin² θ/b² = 1/(x² + y²).

=> (y²/(x² + y²))/a² + (x²/(x² + y²))/b² = 1/(x² + y²)

=> y²/a² + x²/b² = 1

Similar problems

Problem 1: If x = a cos θ − b sin θ and y = a sin θ + b cos θ, then prove that x² + y² = a² + b².

Solution:

x = a cos θ − b sin θ and y = a sin θ + b cos θ

Here, LHS = x² + y²

= (a cos θ − b sin θ)² + (a sin θ + b cos θ)²

= a² cos² θ + b² sin² θ – 2ab cos θ sin θ + a² sin² θ + b² cos² θ + 2ab sin θ cos θ

= a² (cos² θ + sin² θ) + b² (cos² θ + sin² θ)

= a² + b²

= RHS

Hence proved.

Problem 2: If x = a sin θ + b cos θ and y = a cos θ − b sin θ, prove that x² + y² = a² + b².

Solution: 

x = a sin θ + b cos θ and y = a cos θ – b sin θ.

Here, LHS = x² + y²

= (a sin θ + b cos θ)² + (a cos θ – b sin θ)²

= a² sin² θ + b² cos² θ + 2ab sin θ cos θ + a² cos² θ + b² sin² θ – 2ab cos θ sin θ

= a² (sin² θ + cos² θ) + b² (cos² θ + sin² θ)

= a² + b²

= RHS

Hence proved.

Problem 3: If x/a cos θ + y/b sin θ = 1 and x/a sin θ – y/b cos θ = 1, prove that x²/a² + y²/b² = 2.

Solution: 

x/a cos θ + y/b sin θ = 1

Squaring both sides, we get

x²/a² cos² θ + y²/b² sin² θ + 2 (x/a) (y/b) cos θ sin θ = 1 ⇢ (1)

Also, we are given,

x/a sin θ – y/b cos θ = 1

Squaring both sides, we get

x²/a² sin² θ + y²/b² cos² θ – 2 (x/a) (y/b) cos θ sin θ = 1 ⇢ (2)

Adding (1) and (2),

x²/a² cos² θ + y²/b² sin² θ + 2 (x/a) (y/b) cos θ sin θ + x²/a² sin² θ + y²/b² cos² θ – 2 (x/a) (y/b) cos θ sin θ = 1 + 1

x²/a² (cos² θ + sin² θ) + y²/b² (sin² θ + cos² θ) = 2

x²/a² + y²/b² = 2

Hence proved.

Problem 4: If x/a sin θ + y/b cos θ = 1 and x/a cos θ – y/b sin θ = 1, prove that x²/a² + y²/b² = 2.

Solution: 

x/a sin θ + y/b cos θ = 1

Squaring both sides, we get

x²/a² sin² θ + y²/b² cos² θ + 2 (x/a) (y/b) sin θ cos θ = 1 ⇢ (1)

Also, 

x/a cos θ – y/b sin θ = 1

Squaring both sides, we get

x²/a² cos² θ + y²/b² sin² θ – 2 (x/a) (y/b) cos θ sin θ = 1 ⇢ (2)

Adding (1) and (2) gives,

x²/a² sin2 θ + y²/b² cos² θ + 2 (x/a) (y/b) cos θ sin θ + x²/a² cos² θ + y²/b² sin² θ – 2 (x/a) (y/b) cos θ sin θ = 1 + 1

x²/a² (sin² θ + cos² θ) + y²/b² (cos² θ + sin² θ) = 2

x²/a² + y²/b² = 2

Hence proved.

Problem 5: If x = a cos² θ sin θ and y = a sin² θ cos θ, then prove that (x² + y²)³/(x² y²) = a².

Solution:

x = a cos² θ sin θ and y = a sin² θ cos θ

Here, LHS = (x² + y²)³/(x² y²)

= (a² cos⁴ θ sin² θ + a² sin⁴ θ cos² θ)³/[(a² cos⁴ θ sin² θ) (a² sin⁴ θ cos² θ)]

= (a² cos² θ sin² θ (sin² θ + cos² θ))³/(a⁴ cos⁶ θ sin⁶ θ)

= (a² cos² θ sin² θ)³/(a⁴ cos⁶ θ sin⁶ θ)

= (a⁶ cos⁶ θ sin⁶ θ)/(a⁴ cos⁶ θ sin⁶ θ)

= a²

Hence proved.

Problem 6: Prove that (1 – sin θ)/(1 + sin θ) = (sec θ – tan θ)².

Solution:

LHS = (1 – sin θ)/(1 + sin θ)

= ((1 – sin θ) (1 – sin θ))/((1 + sin θ) (1 – sin θ))

= (1 – sin θ)²/(1 – sin² θ)

= (1 – sin θ)²/(cos² θ)

= ((1 – sin θ)/(cos θ))²

= (sec θ – tan θ)²

= RHS

Hence proved.

Problem 7: Prove that: cos θ/(1 – tan θ) – sin θ/(1 – cot θ) = 1/(cos θ – sin θ).

Solution:

LHS = cos θ/(1 – tan θ) – sin θ/(1 – cot θ)

= cos θ/(1 – sin θ/cos θ) – sin θ/(1 – cos θ/sin θ)

= cos² θ/(cos θ – sin θ) + sin² θ (cos θ – sin θ)

= (cos² θ + sin² θ)/(cos θ – sin θ)

= 1/(cos θ – sin θ)

= RHS

Hence proved.