Solution:
We have been given that πππ (Ξ± + Ξ²) = 0.
We know that πππ 90Β° = 0
β πππ (Ξ± + Ξ²) = πππ 90Β°
β Ξ± + Ξ² = 90Β°
β Ξ± = 90Β° β Ξ²
Substituting the value of Ξ± from the above equation, we get
Now, π ππ (Ξ± β Ξ²) = π ππ (90Β° β Ξ² β Ξ²)
β π ππ (Ξ± β Ξ²) = π ππ (90Β° β 2Ξ²)
We know that, for any angle ΞΈ, π ππ (90Β° β ΞΈ) = πππ ΞΈ
Therefore we can say that, π ππ (90Β° β 2Ξ²) = πππ 2Ξ²
Hence, here π ππ (Ξ± β Ξ²) can be reduced to πππ 2Ξ².