Question

cotx = -3/5,-pi/2<x<0

Use identities to find the exact value for the remaining five trigonometric functions

Answer 1

We know that cotx = -3/5.

Since cotx = cosx/sinx, we can solve for sinx and cosx using the Pythagorean identity, sin^2x + cos^2x = 1:

cotx = cosx/sinx

-3/5 = cosx/sinx

Squaring both sides:

9/25 = cos^2x/sin^2x

Using the Pythagorean identity, we know that sin^2x + cos^2x = 1, so we can substitute cos^2x for 1 - sin^2x:

9/25 = (1 - sin^2x)/sin^2x

Multiplying both sides by sin^2x:

9 = 25 - 25sin^2x

25sin^2x = 16

sinx = +/- 4/5

If sinx = 4/5, then cosx = -3/5 (since cotx is negative in the given interval).

Now, we can find the remaining trigonometric functions:

tanx = sinx/cosx = (4/5)/(-3/5) = -4/3

cscx = 1/sinx = 5/4

secx = 1/cosx = -5/3

cotx = cosx/sinx = (-3/5)/(4/5) = -3/4

Therefore, the exact values of the six trigonometric functions are:

sinx = 4/5

cosx = -3/5

tanx = -4/3

cscx = 5/4

secx = -5/3

cotx = -3/4

Answer 2

???? How can you say that cotx = -3/4, when you started out given that cotx = -3/5?

Draw a triangle in QIV with
x = 3
y = -5
so r = √34
sinθ = y/r = -5/√34
cosθ = x/r = 3/√34
tanθ = y/x = -5/3
cotθ = x/y = -3/5
secθ = r/x = √34/3
cscθ = r/y = -√34/5

Answer 3

I apologize for the mistake in my previous response. As you correctly pointed out, cotx = -3/5, and not -3/4. Here are the correct values for the remaining trigonometric functions:

sinx = -4/5 (since sinx is negative in the given interval)

cosx = -3/5

tanx = sinx/cosx = (-4/5)/(-3/5) = 4/3

cscx = 1/sinx = -5/4

secx = 1/cosx = -5/3

cotx = cosx/sinx = (-3/5)/(-4/5) = 3/4

Again, I apologize for any confusion my previous response may have caused. I appreciate you pointing out the error.