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Please solved this..... If sinA +cosA =2 then find the value of sinA -cosA
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Please solved this..... If sinA +cosA =2 then find the value of sinA ...
Solution:

Given, sinA cosA = 2

We know that sin2A = 2sinA cosA

=> sin2A = 4

=> sinA = ±√(4/2) = ±√2

Now, we need to find the value of sinA - cosA

Let’s assume, sinA - cosA = x

Squaring both sides, we get:

(sinA - cosA)^2 = x^2

=> sin^2A + cos^2A - 2sinA cosA = x^2

As sin^2A + cos^2A = 1, we can write:

1 - 2sinA cosA = x^2

As we know sinA cosA = 2, we can substitute it in the above equation:

1 - 2(2) = x^2

=> x^2 = -3

As the value of x^2 is negative, there is no real solution for x.

Therefore, the value of sinA - cosA does not exist.

Answer: The value of sinA - cosA does not exist.
Community Answer
Please solved this..... If sinA +cosA =2 then find the value of sinA ...
sinA + cosA = 2 squaring both side,(sinA + cosA )2= 4sin2A + cos2a + 2sinAcosA = 4NOW sin2A + cos2A = 1therefore,, 2sinAcosA = 4-1that is ..2sinAcosA = 3NOW (sinA - cosA)2 = sin2A - cos2A + 2sinAcosA(sinA-cosA)2 = 4sinA-cosA = 2this the right method
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