# If x sin θ – y cos θ = √(x^{2} + y^{2}) and cos^{2} θ/a^{2} + sin^{2} θ/b^{2} = 1/(x^{2} + y^{2}), then prove that y^{2}/a^{2} + x^{2}/b^{2} = 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.

sin^{2}θ + cos^{2}θ = 1

**Derivation**

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

Applying Pythagoras theorem on this triangle, we get

AC

^{2}= AB^{2}+ BC^{2 }Dividing both sides by AC

^{2}, we getAC

^{2}/AC^{2}= AB^{2}/AC^{2}+ BC^{2}/AC^{2}(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),

sin^{2}θ + cos^{2}θ = 1This proves the identity of sine and cosine ratios.

### If x sin θ – y cos θ = √(x^{2} + y^{2}) and cos^{2} θ/a^{2} + sin^{2} θ/b^{2} = 1/(x^{2} + y^{2}), then prove that y^{2}/a^{2} + x^{2}/b^{2} = 1.

**Solution:**

x sin θ – y cos θ = √(x

^{2}+ y^{2})Squaring both sides we get,

=> x

^{2}sin^{2}θ + y^{2}cos^{2}θ – 2xy sin θ cos θ = x^{2}+ y^{2}=> x

^{2}sin^{2}θ + y^{2}cos^{2}θ – 2xy sin θ cos θ – x^{2}– y^{2}= 0=> x

^{2}(sin^{2}θ – 1) + y^{2}(cos^{2}θ -1) – 2xy sin θ cos θ = 0=> – x

^{2}cos^{2}θ – y^{2}sin^{2}θ – 2xy sin θ cos θ = 0=> x

^{2}cos^{2}θ + y^{2}sin^{2}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

=> x

^{2}/sin^{2}θ = y^{2}/cos^{2}θ=> tan

^{2}θ = x^{2}/y^{2}=> sin

^{2}θ = x^{2}/(x^{2}+ y^{2}) and cos^{2}θ = y^{2}/(x^{2}+ y^{2})Put the above values in the equation cos

^{2}θ/a^{2}+ sin^{2}θ/b^{2}= 1/(x^{2}+ y^{2}).=> (y

^{2}/(x^{2}+ y^{2}))/a^{2}+ (x^{2}/(x^{2}+ y^{2}))/b^{2}= 1/(x^{2}+ y^{2})=> y

^{2}/a^{2}+ x^{2}/b^{2}= 1

### Similar problems

**Problem 1: If x = a cos θ − b sin θ and y = a sin θ + b cos θ, then prove that x ^{2} + y^{2} = a^{2} + b^{2}.**

**Solution:**

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

Here, LHS = x

^{2}+ y^{2}= (a cos θ − b sin θ)

^{2}+ (a sin θ + b cos θ)^{2}= a

^{2}cos^{2}θ + b^{2}sin^{2}θ – 2ab cos θ sin θ + a^{2}sin^{2}θ + b^{2}cos^{2}θ + 2ab sin θ cos θ= a

^{2}(cos^{2}θ + sin^{2}θ) + b^{2}(cos^{2}θ + sin^{2}θ)= a

^{2}+ b^{2}= RHS

Hence proved.

**Problem 2: If x = a sin θ + b cos θ and y = a cos θ − b sin θ, prove that x ^{2} + y^{2} = a^{2} + b^{2}.**

**Solution: **

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

Here, LHS = x

^{2}+ y^{2}= (a sin θ + b cos θ)

^{2}+ (a cos θ – b sin θ)^{2}= a

^{2}sin^{2}θ + b^{2}cos^{2}θ + 2ab sin θ cos θ + a^{2}cos^{2}θ + b^{2}sin^{2}θ – 2ab cos θ sin θ= a

^{2}(sin^{2}θ + cos^{2}θ) + b^{2}(cos^{2}θ + sin^{2}θ)= a

^{2}+ b^{2}= 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 x ^{2}/a^{2} + y^{2}/b^{2} = 2.**

**Solution: **

x/a cos θ + y/b sin θ = 1

Squaring both sides, we get

x

^{2}/a^{2}cos^{2}θ + y^{2}/b^{2}sin^{2}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

x

^{2}/a^{2}sin^{2}θ + y^{2}/b^{2}cos^{2}θ – 2 (x/a) (y/b) cos θ sin θ = 1 ⇢ (2)Adding (1) and (2),

x

^{2}/a^{2}cos^{2}θ + y^{2}/b^{2}sin^{2}θ + 2 (x/a) (y/b) cos θ sin θ + x^{2}/a^{2}sin^{2}θ + y^{2}/b^{2}cos^{2}θ – 2 (x/a) (y/b) cos θ sin θ = 1 + 1x

^{2}/a^{2}(cos^{2}θ + sin^{2}θ) + y^{2}/b^{2}(sin^{2}θ + cos^{2}θ) = 2x

^{2}/a^{2}+ y^{2}/b^{2}= 2Hence 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 x ^{2}/a^{2} + y^{2}/b^{2} = 2.**

**Solution: **

x/a sin θ + y/b cos θ = 1

Squaring both sides, we get

x

^{2}/a^{2}sin^{2}θ + y^{2}/b^{2}cos^{2}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

x

^{2}/a^{2}cos^{2}θ + y^{2}/b^{2}sin^{2}θ – 2 (x/a) (y/b) cos θ sin θ = 1 ⇢ (2)Adding (1) and (2) gives,

x

^{2}/a^{2}sin2 θ + y^{2}/b^{2}cos^{2}θ + 2 (x/a) (y/b) cos θ sin θ + x^{2}/a^{2}cos^{2}θ + y^{2}/b^{2}sin^{2}θ – 2 (x/a) (y/b) cos θ sin θ = 1 + 1x

^{2}/a^{2}(sin^{2}θ + cos^{2}θ) + y^{2}/b^{2}(cos^{2}θ + sin^{2}θ) = 2x

^{2}/a^{2}+ y^{2}/b^{2}= 2Hence proved.

**Problem 5: If x = a cos ^{2} θ sin θ and y = a sin^{2} θ cos θ, then prove that (x^{2} + y^{2})^{3}/(x^{2 }y^{2}) = a^{2}.**

**Solution:**

x = a cos

^{2 }θ sin θ and y = a sin^{2}θ cos θHere, LHS = (x

^{2}+ y^{2})^{3}/(x^{2}y^{2})= (a

^{2}cos^{4}θ sin^{2}θ + a^{2}sin^{4}θ cos^{2}θ)^{3}/[(a^{2}cos^{4}θ sin^{2}θ) (a^{2}sin^{4}θ cos^{2}θ)]= (a

^{2}cos^{2}θ sin^{2}θ (sin^{2}θ + cos^{2}θ))^{3}/(a^{4}cos^{6}θ sin^{6}θ)= (a

^{2}cos^{2}θ sin^{2}θ)^{3}/(a^{4}cos^{6}θ sin^{6}θ)= (a

^{6}cos^{6}θ sin^{6}θ)/(a^{4}cos^{6}θ sin^{6}θ)= a

^{2}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 – sin^{2}θ)= (1 – sin θ)

^{2}/(cos^{2}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)

= cos

^{2}θ/(cos θ – sin θ) + sin^{2}θ (cos θ – sin θ)= (cos

^{2}θ + sin^{2}θ)/(cos θ – sin θ)= 1/(cos θ – sin θ)

= RHS

Hence proved.