If cos A+ sin A =√2cos A; show that cos A - sin A= √2 sin A?

Question

Answer

CosA+SinA=√2CosA,--(given),
or, CosA/CosA+SinA/CosA=√2,
1+TanA=√2,
TanA=√2-1,--(1).
To show --> CosA-SinA=√2 SinA ,
or, CosA/SinA-sin A/SinA=√2,
CotA=√2+1,---(2).
from (1).and (2).
LHS=CotA=√2+1,
=1/TanA ,(CotA=1/TanA),
=1/√2-1,
=1(√2+1)/(√2-1)(√2+1),
=√2+1/2-1,
=√2+1/1,
=√2+1
so, LHS=RHS=√2+1,
Hence proved

Answer

Answer

**Proof:**

Given: cos A sin A = √2 cos A

To prove: cos A - sin A = √2 sin A

Let's start by simplifying the given equation:

cos A sin A = √2 cos A

Dividing both sides by cos A (assuming cos A is not zero):

sin A = √2

Now, let's square both sides of the equation to eliminate the square root:

sin^2 A = (√2)^2

sin^2 A = 2

Next, we can use the trigonometric identity sin^2 A + cos^2 A = 1:

1 - cos^2 A = 2

Rearranging the equation:

cos^2 A = -1

Since the square of any real number cannot be negative, this equation implies that cos A is not a real number. Therefore, the original equation cos A sin A = √2 cos A cannot hold true for any real value of A.

Hence, the given equation is not correct, and we cannot prove cos A - sin A = √2 sin A using the given information.

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