SOLUTION: Verify the identity. cotangent of x to the second power divided by quantity cosecant of x plus one equals quantity one minus sine of x divided by sine of x cot^2(x)/[csc

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Verify the identity. cotangent of x to the second power divided by quantity cosecant of x plus one equals quantity one minus sine of x divided by sine of x cot^2(x)/[csc      Log On


   



Question 1106832: Verify the identity.
cotangent of x to the second power divided by quantity cosecant of x plus one equals
quantity one minus sine of x divided by sine of x

cot^2(x)/[csc(x)+1] = [1 - sin(x)]/sin(x)

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Notice that you didn't include enough parentheses in the way you
typed the problem.  You had THIS:

cot^2 x/csc x+1 = 1 - sin x/ sin x

When you have to type mathematics all on one line, there must be 
enough parentheses to tell where numerators and denominators start 
and end.  You should have typed THIS instead:

cot^2(x)/[csc(x)+1] = [1 - sin(x)]/sin(x)

The way you typed it, there is no way to tell whether the 1's are
part of the denominator on the left and the numerator on the right
or whether they are separate terms.  So learn to be careful to use
enough parentheses when typing mathematics all on one line.  I was
able to tell that you meant because you used the word "quantity".

cot%5E2%28x%29%2F%28csc%5E%22%22%28x%29%2B1%29%22%22=%22%22%281+-+sin%5E%22%22%28x%29%29%2Fsin%5E%22%22%28x%29

Use the identity 1%2Bcot%5E2%28theta%29%22%22=%22%22csc%5E2%28theta%29 solved for cot%5E2%28theta%29%22%22=%22%22csc%5E2%28theta%29-1 
to substitute in the left side for cot%5E2%28x%29.

%28csc%5E2%28x%29-1%29%2F%28csc%5E%22%22%28x%29%2B1%29

Factor the numerator as the difference of two squares:



Cancel:



csc%5E%22%22%28x%29-1

Use the identity csc%5E%22%22%28theta%29%22%22=%22%221%2Fsin%28theta%29

1%5E%22%22%2Fsin%5E%22%22%28x%29-1

Write 1 as 1%2F1

1%5E%22%22%2Fsin%5E%22%22%28x%29-1%5E%22%22%2F1%5E%22%22

Get LCD by multiplying 1%2F1 by sin%28x%29%2Fsin%28x%29



1%5E%22%22%2Fsin%5E%22%22%28x%29-expr%28sin%5E%22%22%28x%29%2Fsin%5E%22%22%28x%29%29

%281+-+sin%5E%22%22%28x%29%29%2Fsin%5E%22%22%28x%29

Edwin