Question

- 1. If Sin (A + B) = 1 and Cos(A - B) = 1, then the value of A and B are respectively equals to * O 30°, 60° , O 40°, 50° O 45°, 45° O None of these. ​

Answer 1

$\textsf{\large{\underline{Solution}:}}$

Given That:

→ sin(A + B) = 1

→ cos(A - B) = 1

We know that:

→ sin(90°) = 1

→ sin(A + B) = sin(90°)

→ A + B = 90°

Again:

→ cos(0°) = 1

→ cos(A - B) = cos(0°)

→ A - B = 0°

→ A = B

Therefore:

→ A + B = 90°

→ A = B = 90°/2

→ A = B = 45°

★ Which is our required answer.

$\textsf{\large{\underline{Learn More}:}}$

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.

• sin²x + cos²x = 1
• cosec²x - cot²x = 1
• sec²x - tan²x = 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

Given That :

➸ Sin (A+B) = 1

➸ Cos (A-B) = 1

To Find :

Values of A,B

Solution :

as we know that,

Sin 90° = 1 & cos 0° = 1

by solving,

⇒ Sin (A + B) = Sin 90°

⇒ A + B = 90° ____eqⁿ(1)

and,

⇒ Cos (A - B) = Cos 0°

⇒ A - B = 0

⇒ A = B ____eqⁿ(2)

By Substituting eq(2) in eq(1). we get ,

Values of

• A = 45°
• B = 45°