Simplify (tan^2θcsc^2θ-1)/(tan^2θ)
This question totally stumps me. I know that csc^2=1/(sin^2) and tan^2=(sin^2)/(cos^2), but I don't see how I can use these identities to simplify the question. What am I missing?
Help is much appreciated!
separate it into 2 fractions
(tan^2θcsc^2θ-1)/(tan^2θ)
= (tan^2θcsc^2θ)/(tan^2θ) - 1/tan^2θ
= csc^2θ - cot^2θ
= 1/sin2θ - cos2θ/sin2θ
= (1 - cos2θ)/sin2θ
= sin2θ/sin2θ
= 1
the part starting from
= 1/sin2θ - cos2θ/sin2θ
= (1 - cos2θ)/sin2θ
= sin2θ/sin2θ
= 1
should say
= 1/sin^2θ - cos^2θ/sin^2θ
= (1 - cos^2θ)/sin^2θ
= sin^2θ/sin^2θ
= 1
I was "cut-and-pasting" and left out the ^
thank you so much for the explanation!
To simplify the given expression, let's start by rewriting the expression using the identities you mentioned.
We have: (tan^2θcsc^2θ - 1) / (tan^2θ)
Using the identity csc^2θ = 1 / sin^2θ, we can replace csc^2θ in the numerator:
= (tan^2θ * 1/sin^2θ - 1) / (tan^2θ)
Now, let's work on simplifying the numerator. Multiplying the fractions, we get:
= [tan^2θ * 1 - 1 * sin^2θ] / (tan^2θ * sin^2θ)
= (tan^2θ - sin^2θ) / (tan^2θ * sin^2θ)
Using the identity tan^2θ = sin^2θ / cos^2θ, we can replace tan^2θ:
= [(sin^2θ / cos^2θ) - sin^2θ] / (sin^2θ * cos^2θ)
To combine the fractions in the numerator, we need a common denominator, which is cos^2θ. So, let's rewrite the numerator:
= (sin^2θ - sin^2θ * cos^2θ) / (sin^2θ * cos^2θ)
Factoring out sin^2θ from the numerator, we get:
= sin^2θ * (1 - cos^2θ) / (sin^2θ * cos^2θ)
Using the identity 1 - cos^2θ = sin^2θ, we can simplify the numerator further:
= sin^2θ * sin^2θ / (sin^2θ * cos^2θ)
= (sin^2θ)^2 / (sin^2θ * cos^2θ)
Now, we can cancel out the common factors in the numerator and denominator:
= sin^2θ / cos^2θ
= tan^2θ
So, the simplified expression is equal to tan^2θ.
In summary, by applying trigonometric identities and simplification techniques, we were able to simplify (tan^2θcsc^2θ - 1) / (tan^2θ) to tan^2θ.