Question

prove that: (sinx+cosx )2 + (sinx-cosx)2=2​

Answer 1

Solution:

We have to prove:

→ (sin θ + cos θ)² + (sin θ - cos θ)² = 2

Taking Left Hand Side, we get:

= (sin θ + cos θ)² + (sin θ - cos θ)²

We know that:

→ (a + b)² = a² + 2ab + b²

→ (a - b)² = a² - 2ab + b²

Therefore, LHS becomes:

= (sin²θ + 2 sin θ cos θ + cos²θ) + (sin²θ - 2 sin θ cos θ + cos²θ)

= sin²θ + cos²θ + sin²θ + cos²θ

= 2(sin²θ + cos²θ)

= 2 × 1

= 2

Therefore:

→ (sin θ + cos θ)² + (sin θ - cos θ)² = 2

Hence Proved..!!

To Know More:

1. Relationship between sides and T-Ratios.

• sin θ = Height/Hypotenuse
• cos θ = Base/Hypotenuse
• tan θ = Height/Base
• cot θ = Base/Height
• sec θ = Hypotenuse/Base
• cosec θ = Hypotenuse/Height

2. Square formulae.

• sin²θ + cos²θ = 1
• cosec²θ - cot²θ = 1
• sec²θ - tan²θ = 1

3. Reciprocal Relationship.

• sin θ = 1/cosec θ
• cos θ = 1/sec θ
• tan θ = 1/cot θ
• cosec θ = 1/sin θ
• sec θ = 1/cos θ
• tan θ = 1/cot θ

4. Cofunction identities.

• sin(90° - θ) = cos θ
• cos(90° - θ) = sin θ
• cosec(90° - θ) = sec θ
• sec(90° - θ) = cosec θ
• tan(90° - θ) = cot θ
• cot(90° - θ) = tan θ

5. Even odd identities.

• sin -θ = -sin θ
• cos -θ = cos θ
• tan -θ = -tan θ