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