SOLUTION: Find all solutions of the equation tan theta+cot theta+sec theta+csc theta=0

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Question 612288: Find all solutions of the equation tan theta+cot theta+sec theta+csc theta=0
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
tan%28theta%29%2Bcot%28theta%29%2Bsec%28theta%29%2Bcsc%28theta%29=0
When solving equations like this you often start by using algebra and/or Trig properties to transform the equation into one or more equations of the form:
TrigFunction(expression) = number

When it is not clear how to make this transformation, then it can be helpful to use Trig properties to transform all the Trig functions into sin's and/or cos's. Replacing tan with sin/cos, cot with cos/sin, sec with 1/cos and csc with 1/sin we get:


Let's see what happens if we eliminate the fractions. This can be done if we multiply both sides by the lowest common denominator (LCD). The LCD here is sin%28theta%29%2Acos%28theta%29:

Using the Distributive Property we get:

The denominators all cancel leaving:
sin%5E2%28theta%29+%2B+cos%5E2%28theta%29+%2B+sin%28theta%29+%2B+cos%28theta%29+=+0
Since sin%5E2%28theta%29+%2B+cos%5E2%28theta%29+=+1 this simplifies to:
1+%2B+sin%28theta%29+%2B+cos%28theta%29+=+0

Believe it or not, we have made progress. We started with an equation of 4 different Trig functions and we now have an equation of just two. And we can get it down to just one by...
Subtracting cos%28theta%29 we get:
1+%2B+sin%28theta%29+=+-cos%28theta%29
Squaring both sides:
%281+%2B+sin%28theta%29%29%5E2+=+%28-cos%28theta%29%29%5E2
1+%2B+2sin%28theta%29+%2B+sin%5E2%28theta%29+=+cos%5E2%28theta%29
Replacing cos%5E2%28theta%29 with 1-sin%5E2%28theta%29:
1+%2B+2sin%28theta%29+%2B+sin%5E2%28theta%29+=+1-sin%5E2%28theta%29

Now that we have just one function the equation is a quadratic in sin%28theta%29. So we want one side to be zero. Subtracting 1 and adding sin%5E2%28theta%29 we get:
2sin%5E2%28theta%29%2B2sin%28theta%29+=+0
Factoring we get:
2sin%28theta%29%2A%28sin%28theta%29+%2B+1%29+=+0
From the Zero Product Property:
2sin%28theta%29+=+0 or sin%28theta%29+%2B+1+=+0
Solving each of these we get:
sin%28theta%29+=+0 or sin%28theta%29+=+-1

After all this, we have finally transformed
tan%28theta%29%2Bcot%28theta%29%2Bsec%28theta%29%2Bcsc%28theta%29=0
into
sin%28theta%29+=+0 or sin%28theta%29+=+-1

From these we can now write the general solution. Since sin is zero at 0 and 180 degrees:
theta+=+0+%2B+360n
or
theta+=+180+%2B+360n
And since sin is -1 at 270:
theta+=+270+%2B+360n

Last of all, we need to check. This is not optional! During the course of our transformations we
  • Multiplied both sides of the equation by an expression that could be zero when we multiplied both sides by sin%28theta%29%2Acos%28theta%29
  • Squared both sides of the equation.
Neither one of these actions is wrong. But they each can introduce what are called extraneous solutions. Extraneous solutions are "solutions" that fit the modified equation but do not fit the original equation. If you do either one these things, then you must check your solutions to make sure they actually work.

Use the original equation to check:
tan%28theta%29%2Bcot%28theta%29%2Bsec%28theta%29%2Bcsc%28theta%29=0
Checking theta+=+0:
tan%280%29%2Bcot%280%29%2Bsec%280%29%2Bcsc%280%29=0
cot and sec are undefined at 0! So we must reject this and all the co-terminal solutions.
Checking theta+=+180:
tan%28180%29%2Bcot%28180%29%2Bsec%28180%29%2Bcsc%28180%29=0
cot and sec are also undefined at 180! So we must reject this and all the co-terminal solutions.
Checking theta+=+270:
tan%28270%29%2Bcot%28270%29%2Bsec%28270%29%2Bcsc%28270%29=0
tan and csc are undefined at 270! So we must reject this and all the co-terminal solutions.

So after all this, we have rejected every solution we found. Therefore your equation has no solution.