SOLUTION: tan theta + cot theta = 1 over the sin theta times cos theta It asks me to: "Verify the Identity. Justify each step."

Algebra ->  Trigonometry-basics -> SOLUTION: tan theta + cot theta = 1 over the sin theta times cos theta It asks me to: "Verify the Identity. Justify each step."      Log On


   



Question 31257: tan theta + cot theta = 1 over the sin theta times cos theta
It asks me to:
"Verify the Identity. Justify each step."

Found 3 solutions by longjonsilver, chibisan, jsmallt9:
Answer by longjonsilver(2297) About Me  (Show Source):
You can put this solution on YOUR website!
+tanA+%2B+cotA+=+1%2F%28sinAcosA%29+

i shall work on the left hand side, since this contains items that are "more complicated". It is easier to start with a complicated expression and simplify it rather than work the other way:

+tanA+%2B+cotA+
+%28%28sinA%29%2F%28cosA%29%29+%2B+%28%28cosA%29%2F%28sinA%29%29+

+%28%28sinA%29%5E2%2F%28sinAcosA%29%29+%2B+%28%28cosA%29%5E2%2F%28sinAcosA%29%29+
+%28%28%28sinA%29%5E2+%2B+%28cosA%29%5E2%29%29%2F%28sinAcosA%29+
+1%2F%28sinAcosA%29+

jon.

Answer by chibisan(131) About Me  (Show Source):
You can put this solution on YOUR website!
tanθ + cotθ = 1/sinθcosθ
L.H.S
sinθ/cosθ + cosθ/sinθ
sin^2θ+cos^2θ/sinθcosθ
1/sinθcosθ

Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
(Since Algebra.com's formula software does not "do" theta for some reason, I'll use x, instead).

As a general rule, when you are trying to solve a Trig. problem and you don't see any better way, express everything in terms of sin and cos. This will usually result in something you can figure out.

Using this on your expression:
tan%28x%29+%2B+cot%28x%29+=+1%2F%28sin%28x%29%2Acos%28x%29%29
Replacing tan(x) with sin(x)/cos(x) and cot(x) with cos(x)/sin(x):
sin%28x%29%2Fcos%28x%29+%2B+cos%28x%29%2Fsin%28x%29+=+1%2F%28sin%28x%29%2Acos%28x%29%29
Verifying identities often involves changing the equation, using proper Trig. and/or Algebra, so that the two sides match. So we will try to match the left and right sides. Since the right side is a single fraction, we'll go ahead and add the two fractions on the left tomake the left side one fraction. And to add fractions we nedd, of course, the same denominator:

Adding we get:

Since the numerator is part of a well known Trig. identity: %28sin%28x%29%29%5E2+%2B+%28cos%28x%29%29%5E2+=+1, we get:
1%2F%28sin%28x%29%2Acos%28x%29%29+=+1%2F%28sin%28x%29%2Acos%28x%29%29
And we have matched the left and right sides which verifies the identity!