Question

sin (60 - A) cos (30 + A) + cos (60 - A) sin (30 + A) = 1​

Answer 1

Solution:

We have to prove that:

→ sin(60° - A) cos(30° + A) + cos(60° - A) sin(30° + A) = 1

Taking Left Hand Side, we get:

= sin(60° - A) cos(30° + A) + cos(60° - A) sin(30° + A)

Can be written as:

= sin(90° - (30° + A)) cos (30° + A) + cos(90° - (30° + A)) sin(30° + A)

We know that:

→ sin(90° - x) = cos(x)

→ cos(90° - x) = sin(x)

Therefore, LHS becomes:

= cos(30° + A) cos(30° + A) + sin(30° + A) sin(30° + A)

= cos²(30° + A) + sin²(30° + A)

We know that:

→ sin²(x) + cos²(x) = 1

Therefore, LHS becomes:

= 1

Therefore:

→ sin(60° - A) cos(30° + A) + cos(60° - A) sin(30° + A) = 1

Hence Proved..!!

Additional Information:

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