Question

Can someone show me how to do this, please?

Verify the Pythagorean identity.

Answer 1

\[1+\cot^2 \theta = \csc^2 \theta\]

Answer 2

Add the two terms on the left.
Write the cotangent as a fraction, find a common denominator, and add them together.
Then pay close attention to the numerator of the sum.

Answer 3

Umm, do you think you could simplify that at all?

Answer 4

Here are all the steps in one post


1+cot2(θ)=csc2(θ)


1+cos2(θ)sin2(θ)=csc2(θ)


sin2(θ)sin2(θ)+cos2(θ)sin2(θ)=csc2(θ)


sin2(θ)+cos2(θ)sin2(θ)=csc2(θ)


1sin2(θ)=csc2(θ)


csc2(θ)=csc2(θ)


the two sides of the last equation are identical, so we have confirmed the identity

Answer 5

\(1+\cot^2 \theta = \csc^2 \theta\)

\(1 + \dfrac{\cos^2 \theta}{\sin^2 \theta} = \csc^2 \theta\)

We need a common denominator to add the left side.

\(\dfrac{\sin^2 \theta}{\sin^2 \theta}+ \dfrac{cos^2 \theta}{\sin^2 \theta} = \csc^2 \theta\)

\(\dfrac{\sin^2 \theta + cos^2 \theta}{\sin^2 \theta} = \csc^2 \theta\)

Use the identity: \(\sin^2 \theta + \cos^2 \theta = 1\)

\(\dfrac{1}{\sin^2 \theta} = \csc^2 \theta\)

\(\csc^2 \theta = \csc^2 \theta\)

Answer 6

Awesome! Thanks a bunch for explaining it

Answer 7

You're welcome.