Plszee answer this question and make ur brilliant answer.
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1
Hola there,
Let theta be 'A'
Given => cosA + sinA = √2 cosA
To prove => cosA - sinA = √2 sinA
Proof => cosA + sinA = √2 cosA
Squaring the above equation..
=> (cosA + sinA = √2 cosA)²
=> cos²A + sin²A + 2sinAcosA = 2cos²A
=> 1 - sin²A + 1 - cos²A + 2sinAcosA = 2cos²A
=> 2 - 2cos²A = sin²A + cos ²A - 2sinAcosA
=> 2(1 - cos²A) = (cosA - sinA)²
=> 2sin²A = (cosA - sinA)²
=> cosA - sinA = √2sin²A
=> cosA - sinA = √2 sinA
Hence Proved.
Hope this helps...:)
Let theta be 'A'
Given => cosA + sinA = √2 cosA
To prove => cosA - sinA = √2 sinA
Proof => cosA + sinA = √2 cosA
Squaring the above equation..
=> (cosA + sinA = √2 cosA)²
=> cos²A + sin²A + 2sinAcosA = 2cos²A
=> 1 - sin²A + 1 - cos²A + 2sinAcosA = 2cos²A
=> 2 - 2cos²A = sin²A + cos ²A - 2sinAcosA
=> 2(1 - cos²A) = (cosA - sinA)²
=> 2sin²A = (cosA - sinA)²
=> cosA - sinA = √2sin²A
=> cosA - sinA = √2 sinA
Hence Proved.
Hope this helps...:)
Frnd es maine much wrong hai?
can u tell which step is wrong?
Sorry it is fully correct
then why did u said it's wrong
I suggest u to please atleast see the answer before saying anyone wrong..
Answered by
1
this is your answer
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