SOLUTION: Simplify [1/(tanx + cotx + secx + cosecx)]²

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Question 1208297: Simplify
[1/(tanx + cotx + secx + cosecx)]²

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!

%281%2F%28tan%28x%29+%2B+cot%28x%29+%2B+sec%28x%29+%2B+csc%28x%29%29%29%5E2 

1%5E%22%22%2F%28tan%28x%29+%2B+cot%28x%29+%2B+sec%28x%29+%2B+csc%28x%29%5E%22%22%29%5E2

Draw a right triangle with angle x, hypotenuse 1, and legs "a" and "b"


 
1%5E%22%22%2F%28a%2Fb+%2B+b%2Fa+%2B+1%2Fb+%2B+1%2Fa%29%5E2

Since , that's
because by the Pythagorean theorem, a2+b2=1

1%5E%22%22%2F%28%281%2Ba%2Bb%29%2F%28ab%29%29%5E2

Multiply top and bottom by a2b2

a%5E2b%5E2%2F%281%2Ba%2Bb%29%5E2

We multiply out the denominator,

a%5E2b%5E2%2F%281%2Ba%5E2%2Bb%5E2%2B2a%2B2b%2B2ab%29

Since a2+b2=1,

a%5E2b%5E2%2F%282%2B2a%2B2b%2B2ab%29

expr%281%5E%22%22%2F2%5E%22%22%29%2A%28a%5E2b%5E2%2F%281%2Ba%2Bb%2Bab%5E%22%22%29%29

The denominator factors

expr%281%5E%22%22%2F2%5E%22%22%29%2A%28a%5E2b%5E2%2F%28%281%2Ba%29%281%2Bb%29%29%29

Multiply by the product of the "conjugates" of the
factors on the bottom over itself. (Sort of like
rationalizing the denominator)


expr%281%5E%22%22%2F2%5E%22%22%29%2A%28a%5E2b%5E2%2F%28%281%2Ba%29%281%2Bb%29%29%29%22%22%2A%22%22%28%28%281-a%29%281-b%29%5E%22%22%29%2F%28%281-a%29%281-b%29%5E%22%22%29%29

 
expr%281%2F2%29%2A%28a%5E2b%5E2%281-a%29%281-b%29%29%2F%28%281-a%5E2%29%281-b%5E2%29%29%29

By the Pythagorean theorem  1-a2=b2 and 1-b2=a2

expr%281%2F2%29%2A%28a%5E2b%5E2%281-a%29%281-b%29%29%2F%28b%5E2a%5E2%29%29%29

Canceling:

expr%281%2F2%29%2A%281-a%29%281-b%29

Since a = sin(x) and b = cos(x)

expr%281%2F2%29%2A%281%5E%22%22-sin%28x%29%29%281%5E%22%22-cos%28x%29%29  <--simplest form

Edwin