If x is acute angle, and tan x=3 , evaluate cos x−sin x

_ __________
4 cos x+sin x

Use / as your division sign, enclose numerators and denominators in brackets.

e.g. cos x−sin x
_ __________
4 cos x+sin x

could mean: cosx/4 - sinx/(cosx + sinx)

I will assume that is what you meant.

given tanx = 3 = 3/1 = y/x , so make a sketch of your triangle
and x^2 + y^2 = r^2
1^2 + 3^2 = r^2
r = √10
since x is acute, sinx = 3/√10 , and cosx = 1/√10

cosx/4 - sinx/(cosx + sinx)
= 1/(4√10) - (3/√10) / (4√10)
= 1/(4√10) - 3/4
= (1 - 3√10)/(4√10)

or (√10 - 30)/40 after rationalizing.

To evaluate cos x - sin x / cos x + sin x, we can use trigonometric identities to rewrite the expression in terms of tan x:

cos x - sin x / cos x + sin x = (cos x - sin x) / (cos x + sin x) * (cos x - sin x) / (cos x + sin x)

Now, let's simplify the expression:

( cos x - sin x )^2 / ( cos x + sin x )^2 = ( cos^2 x - 2sin x cos x + sin^2 x ) / ( cos^2 x + 2sin x cos x + sin^2 x )

Using the Pythagorean identity for sine and cosine (sin^2 x + cos^2 x = 1):

(1 - 2sin x cos x)/(1 + 2sin x cos x)

Next, let's use the given information that tan x = 3 to calculate sin x and cos x.

We know that tan x = sin x / cos x, so sin x = 3 cos x.

Substituting this into our expression, we have:

(1 - 2(3cos x)cos x)/(1 + 2(3cos x)cos x) = (1 - 6cos^2 x)/(1 + 6cos^2 x)

Finally, we can simplify this expression further by factoring out a common factor of -1 in the numerator:

-(6cos^2 x - 1)/(1 + 6cos^2 x)

So, the expression cos x - sin x / cos x + sin x is equal to -(6cos^2 x - 1)/(1 + 6cos^2 x).