Question

Value is cos theta + sin theta bracket square + cos theta minus sin theta (square

Answer 1

Solution:

We have to simplify the given expression.

= (cos θ + sin θ)² + (cos θ - sin θ)²

We know that:

→ (a + b)² = a² + 2ab + b²

→ (a - b)² = a² - 2ab + b²

Therefore the expression becomes:

= cos²θ + sin²θ + 2 sinθ cosθ + cos²θ + sin²θ - 2 sinθ cosθ

= 2(cos²θ + sin²θ)

= 2 × 1 [As cos²θ + sin²θ = 1]

= 2

★ Which is our required answer.

Learn More:

1. Relationship between sides and T-Ratios.

• sin(x) = Height/Hypotenuse
• cos(x) = Base/Hypotenuse
• tan(x) = Height/Base
• cot(x) = Base/Height
• sec(x) = Hypotenuse/Base
• cosec(x) = Hypotenuse/Height

2. Square formulae.

• sin²(x) + cos²(x) = 1
• cosec²(x) - cot²(x) = 1
• sec²(x) - tan²(x) = 1

3. Reciprocal Relationship.

• sin(x) = 1/cosec(x)
• cos(x) = 1/sec(x)
• tan(x) = 1/cot(x)

4. Cofunction identities.

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

5. Even odd identities.

• sin(-x) = -sin(x)
• cos(-x) = cos(x)
• tan(-x) = -tan(x)

Answer 2

Answer:

$\rm 2$

Step-by-step explanation:

We have,

$\rm ( \cos θ + \sin θ)^2 + ( \cos θ - \sin θ)^2$

We know that,

$\boxed{(a+b)^2 = a^2 +b^2 +2ab}$

$\boxed{(a-b)^2 = a^2 +b^2 -2ab}$

$\rm \cos ^2 θ + \sin ^2 θ +2 \cos θ \sin θ + \cos ^2 θ + \sin ^2 θ - 2 \cos θ \sin θ$

We know that,

$\boxed{ \cos ^2 θ + \sin ^2 θ =1 }$

$\rm 1 +2 \cos θ \sin θ + 1 - 2 \cos θ \sin θ$

$\rm 1 +\cancel{2 \cos θ \sin θ} + 1 - \cancel{2 \cos θ \sin θ}$

$\rm 1+1 = 2$

$\rm Hence, \: our \: answer \: is \: 2$