Question

@dude @jhonyy9 there's one more that idk of, healp please

Answer 1

prove the left equals the right side

Answer 2

@Angle
@Tranquility

Answer 3

This one seems relatively straightforward. If you look at both sides, what's the first thing you notice?

We have something like:

a^4 - b^4 = a^2 + b^2

That means if we factor the left side, we're going to get some identity that makes (a^2 - b^2) = 1 so that we're left with the RHS

Can you factor
\[\csc^4 x - cot^4 x\]

Answer 4

If you need a little push in the right direction:
\[ a^2 - b^2 = (a+b)(a-b)\]

Answer 5

[(csc x - cot x)(cot x + cot x)]^2

Answer 6

No.

\[ a^4 - b^4 = (a^2)^2 - (b^2)^2\]

Now it's in the form of
\[ x^2 - y^2 = (x + y)(x-y)\]
\[ x = a^2 \\y = b^2\]


What you wrote would end up being
\[(\csc x - \cot x)^2(\cot x + \cot x)^2\]
which is not equivalent to the original expression and is thus wrong

Answer 7

csc^4 x−cot^4 x
(csc^2 - cot^2 x)(csc^2 + cot^2 x)

Answer 8

Now apply one of the Pythagorean identities
\[\cot^2\theta + 1 = \csc^2 \theta\]

Answer 9

Oh
thank youuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu

Answer 10

Anytime!

Answer 11

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Tranquility
Now apply one of the Pythagorean identities
\[\cot^2\theta + 1 = \csc^2 \theta\]
\(\color{#0cbb34}{\text{End of Quote}}\)
explained perfect - congrats !