Question

Question is in attachment Plz solve it Urgent !

More than type answer -,-​

Answer 1

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

Given:

→ p = sin(α - β) sin(γ - δ)

→ q = sin(β - γ) sin(α - δ)

→ r = sin(γ - α) sin(β - δ)

We know that:

→ 2 sin(x) sin(y) = cos(x - y) - cos(x + y)

Therefore:

→ p = sin(α - β) sin(γ - δ)

→ 2p = 2 sin(α - β) sin(γ - δ)

→ 2p = cos(α - β - γ + δ) - cos(α - β + γ - δ) – (i)

Similarly, we can write:

→ 2q = cos(β - γ - α + δ) - cos(β - γ + α - δ) – (ii)

→ 2r = cos(γ - α - β + δ) - cos(γ - α + β - δ) – (iii)

Adding (i), (ii) and (iii), we get:

→ 2p + 2q + 2r = cos(α - β - γ + δ) - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ) - cos(γ - α + β - δ)

We know that:

→ cos(x) = cos(-x)

Therefore:

→ 2p + 2q + 2r = cos(α - β - γ + δ) - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ) - cos(γ - α + β - δ)

→ 2p + 2q + 2r = cos(α - β - γ + δ) - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ) - cos(-(γ - α + β - δ))

→ 2p + 2q + 2r = cos(α - β - γ + δ) - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ) - cos(α - β - γ + δ)

cos(α - β - γ + δ) gets cancelled out. We get:

→ 2p + 2q + 2r = - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ)

Again, applying the same identity, we get:

→ 2p + 2q + 2r = - cos(α - β + γ - δ) + cos(-(β - γ - α + δ)) - cos(β - γ + α - δ) + cos(γ - α - β + δ)

→ 2p + 2q + 2r = - cos(α - β + γ - δ) + cos(α - β + γ - δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ)

cos(α - β + γ - δ) gets cancelled out. We get:

→ 2p + 2q + 2r = - cos(β - γ + α - δ) + cos(γ - α - β + δ)

This too gets cancelled out. So, we get:

→ 2p + 2q + 2r = 0

→ p + q + r = 0

★ Hence, option b is the correct answer for the problem.

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

1. Relationship between sides.

• 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) and vice versa.
• cosec(90° - x) = sec(x) and vice versa.
• tan(90° - x) = cot(x) and vice versa.

5. Even odd identities.

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

Answer 2

Answer:

Solution

:

Given:

→ p = sin(α - β) sin(γ - δ)

→ q = sin(β - γ) sin(α - δ)

→ r = sin(γ - α) sin(β - δ)

We know that:

→ 2 sin(x) sin(y) = cos(x - y) - cos(x + y)

Therefore:

→ p = sin(α - β) sin(γ - δ)

→ 2p = 2 sin(α - β) sin(γ - δ)

→ 2p = cos(α - β - γ + δ) - cos(α - β + γ - δ) – (i)

Similarly, we can write:

→ 2q = cos(β - γ - α + δ) - cos(β - γ + α - δ) – (ii)

→ 2r = cos(γ - α - β + δ) - cos(γ - α + β - δ) – (iii)

Adding (i), (ii) and (iii), we get:

→ 2p + 2q + 2r = cos(α - β - γ + δ) - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ) - cos(γ - α + β - δ)

We know that:

→ cos(x) = cos(-x)

Therefore:

→ 2p + 2q + 2r = cos(α - β - γ + δ) - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ) - cos(γ - α + β - δ)

→ 2p + 2q + 2r = cos(α - β - γ + δ) - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ) - cos(-(γ - α + β - δ))

→ 2p + 2q + 2r = cos(α - β - γ + δ) - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ) - cos(α - β - γ + δ)

cos(α - β - γ + δ) gets cancelled out. We get:

→ 2p + 2q + 2r = - cos(α - β + γ - δ) + cos(β - γ - α + δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ)

Again, applying the same identity, we get:

→ 2p + 2q + 2r = - cos(α - β + γ - δ) + cos(-(β - γ - α + δ)) - cos(β - γ + α - δ) + cos(γ - α - β + δ)

→ 2p + 2q + 2r = - cos(α - β + γ - δ) + cos(α - β + γ - δ) - cos(β - γ + α - δ) + cos(γ - α - β + δ)

cos(α - β + γ - δ) gets cancelled out. We get:

→ 2p + 2q + 2r = - cos(β - γ + α - δ) + cos(γ - α - β + δ)

This too gets cancelled out. So, we get:

→ 2p + 2q + 2r = 0

→ p + q + r = 0

★ Hence, option b is the correct answer for the problem.

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

Learn More

:

1. Relationship between sides.

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) and vice versa.

cosec(90° - x) = sec(x) and vice versa.

tan(90° - x) = cot(x) and vice versa.

5. Even odd identities.

sin(-x) = -sin(x)

cos(-x) = cos(x)

tan(-x) = -tan(x)

hope it's help you