how do u do this problem? cscx+cotx=1?

Question

thanx 4 ur help!!^_^

Answer 1

1/sinx + cosx/sinx = 1
(1 + cosx)/sinx = sinx/sinx
1 + cosx = sinx

Answer 2

csc(x) + cot(x) = 1

Your first step would be to change everything to sines and cosines.

1/sin(x) + cos(x)/sin(x) = 1

Merge the two fractions under a common denominator,

[1 + cos(x)] / sin(x) = 1

Multiply both sides by sin(x)

1 + cos(x) = sin(x)

Bring the cos(x) to the right hand side,

1 = sin(x) - cos(x)

Now, square both sides.

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

Rearrange the terms,

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

Now, note the identity sin^2(x) + cos^2(x) = 1. So we have

1 = 1 - 2sin(x)cos(x)

Subtract 1 both sides,

0 = -2sin(x)cos(x)

Divide -2 both sides,

0 = sin(x)cos(x)

Now, we equate each to 0.

sin(x) = 0
cos(x) = 0

I'm going to assume there's a restricted interval of 0 <= x < 2pi.
For sin(x) = 0, x = {0, pi}
For cos(x) = 0, x = {pi/2, 3pi/2}

Therefore, our (potential) solutions are x = {0, pi, pi/2, 3pi/2}

However, we can't assume all of these values will work, because one of our steps involved squaring both sides of the equation (this does funny things sometimes, since if x = 1, then x^2 = 1, and solving this back, we get x = -1 and 1, so we've just added solutions. That's what we're checking for here; whether we've added solutions). We have to TEST these values by plugging them into our ORIGINAL equation, csc(x) + cot(x) = 1.

Test x = 0. Then
LHS = csc(x) + cot(x). But this isn't defined at x = 0 (since csc(0) is undefined). csc(0) is undefined because
csc(x) = 1/sin(x), and 1/sin(0) means we have 1/0, and that is undefined. Reject this solution.

Test x = pi. By the same reason, we get an undefined answer and reject this solution.

Test x = pi/2: Then
LHS = csc(pi/2) + cot(pi/2) = 1/sin(pi/2) + cos(pi/2) / sin(pi/2)
LHS = 1/1 + 0/1 = 1 = RHS
This solution works.

Test x = 3pi/2. Then
LHS = csc(3pi/2) + cot(3pi/2) = 1/sin(3pi/2) + cos(3pi/2)/sin(3pi/2)
LHS = 1/(-1) + 0/(-1) = -1, which does NOT equal 1.
Reject this solution.

Therefore, our only solution in the interval from [0, 2pi) is x = pi/2.

If we wanted a general solution, it would be

x = {pi/2 + 2kpi | k an integer}

Answer 3

solution

cscx + cotx = 1
1/senx + cosx/senx = 1
(cosx+1)/senx = 1
(cosx+1) = senx <---now we use exp 2 on each member
(cosx+1)^2 = (senx)^2
cos^2(x)+2cosx+1 = sen^2(x)
cos^2(x)+2cosx+1 = 1-cos^2(x)
2cos^2(x)+2cosx = 1-1
2cos^2(x)+2cosx = 0<------we take 2cosx common factor

2cosx(cosx+1) = 0
solutions

cosx = 0 ---> x= pi/2 , 3pi/2
cosx = -1---> x= pi


hope to be useful
nexus