# 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)

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)

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.

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