If πππ  (Ξ± + Ξ²) = 0, then π ππ (Ξ± β Ξ²) can be reduced to

# If πππ  (Ξ± + Ξ²) = 0, then π ππ (Ξ± β Ξ²) can be reduced to

1. A

πππ  Ξ²

2. B

πππ  2Ξ²

3. C

π ππ Ξ±

4. D

π ππ 2Ξ±

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### Solution:

We have been given that πππ  (Ξ± + Ξ²) = 0

We know that πππ  90Β° = 0

β πππ  (Ξ± + Ξ²) = πππ  90Β°

β Ξ± + Ξ² = 90Β°

β Ξ± = 90Β° β Ξ²

Substituting the value of Ξ± from the above equation, we get

Now, π ππ (Ξ± β Ξ²) = π ππ (90Β° β Ξ² β Ξ²)

β π ππ (Ξ± β Ξ²) = π ππ (90Β° β 2Ξ²)

We know that, for any angle ΞΈ, π ππ (90Β° β ΞΈ) = πππ  ΞΈ

Therefore we can say that, π ππ (90Β° β 2Ξ²) = πππ  2Ξ²

Hence, here π ππ (Ξ± β Ξ²) can be reduced to πππ  2Ξ².

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