Question

If sinA-cosA=1, Prove that CosA+SinA=+_1

Answer 1

⇒ sinA - cosA = 1

       Square on both sides:

⇒ (sinA - cosA)² = 1²

⇒ sin²A + cos²A - 2sinAcosA = 1

⇒ 1 - 2sinAcosA = 1

⇒ -2sinAcosA = 0  

       ∴ 2sinAcosA = 0   [also]

Therefore,   using sin²A + cos²A = 1

     Adding 2sinAcosA to both sides:

⇒ sin²A + cos²A + 2sinAcosA = 1 + 2sinAcosA

⇒ (sinA + cosA)² = 1 + 0         [2sinAcosA = 0]

⇒ (sinA + cosA)² = 1

⇒ sinA + cosA =  ± 1

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Answer 2

Given :

• Sin A - Cos A

To Find :

• Prove that Cos A + Sin A = ± 1

Explanation :

• $\sf Sin \: \: A - Cos \: \: A = 1$

Now, Take Square From Both Sides

• $\sf \red{ (Sin \: \: A - Cos \: \: A )² } = \green{ (1)²}$

• $\sf \red{(Sin \: \: A - Cos \: \: A )²} = \green {1 }$

[ We know that, Sin² A + Cos ² A = 1 ]

So, we can put this value

• $\sf \red{(Sin \: \: A - Cos \: \: A )²} = \green{Sin² \: \: A + Cos² \: \: A }$

• $\sf \red{Sin² \: \: A + Cos² \: \: A - 2 × Sin \: \: A × Cos \: \: A } = \green{ Sin² \: \: A + Cos² \: \: A}$

• $\sf \red{ Sin² \: \: A + Cos² \: \: A } = \green{Sin² \: \: A + Cos² \: \: A + 2 × Sin \: \: A × Cos \: \: A }$

• $\sf \red{Sin² \: \: A + Cos² \: \: A} = \green{ (Sin \: \: A + Cos \: \: A )² }$

• $\sf \red{1 } = \green{(Sin \: \: A + Cos \: \: A )²}$

• $\sf \red{ (Sin \: \: A + Cos \: \: A )²} = \green{1}$

• $\sf \red{(Sin \: \: A + Cos \: \: A )} = \green{ ± \sqrt{1}}$

• $\sf \red{Cos \: \: A + Sin \: \: A } = \green{± 1}$

I hope it helps you ❤️✔️