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## Answers

The Brainliest Answer!

**aCosФ + bSinФ=m ; asinФ-bcosФ=n**

**Taking L.H.S,**

**m²+n²**

**=a²Cos²Ф+b²Sin²Ф+2abCosФSinФ+b²Cos²Ф+a²Sin²Ф-2abCosФSinФ**

**Since 2abCosФSinФ is opposite in sign it gets cancelled,**

**=a²Cos²Ф+b²Sin²Ф+a²Sin²Ф+b²Cos²Ф**

**Pairing up the terms,**

**=a²Cos²Ф+a²Sin²Ф+b²Sin²Ф+b²Cos²Ф**

**Now taking a² and b² common,**

**=a²(Cos²Ф+Sin²Ф)+b²(Sin²Ф+Cos²Ф)**

**Since Cos²Ф+Sin²Ф=1**

**=a²+b²(Hence, Proved)**

**L.H.S=R.H.S**

**m²+n²=a²+b²**

asinФ - bcosФ = n

m² + n² = a²cos²Ф + b²cos²Ф + 2absinФcosФ + a²sin²Ф + b²cos²Ф - 2abcosФsinФ

m² + n² = a²(cos²Ф + sin²Ф) + b²(cos²Ф + sin²Ф)

m² + n² = a² + b² prove (cos²Ф + sin²Ф = 1)