Question

Verify csc^4x-cot^4x=2csc^2x-1

Answer 1

First, factor the left side:

$(csc^2(x))^2-(cot^2(x))^2$

This can be simplified using $x^2-y^2 = (x+y)(x-y)$ where the csc term is x and the cot term is y:

$(csc^2(x)+cot^2(x))(csc^2(x)-cot^2(x))$

One of our Pythagorean Identities states that:

$csc^2(x)-cot^2(x)=1$

So the left side is now:

$csc^2(x)+cot^2(x)$

Now we can use the same Pythagorean Identity to change the right side to:

$2csc^2(x)-(csc^2(x)-cot^2(x))$

All we did was express 1 in terms of csc and cot.

Distribute the negative:

$2csc^2(x)-csc^2(x)+cot^2(x)$

Combine like terms:

$csc^2(x)+cot^2(x)$

Now the left and right sides are the same.

Answer 2

Following are the proving to the given expression:

Given:

$\bold{csc^4x-cot^4x=2csc^2x-1}$

To find:

Proving=?

Solution:

$\to \bold{csc^4x-cot^4x=2csc^2x-1}$

Using formula:

$\to (a^2-b^2)=(a-b)(a+b)$

Solving the L.H.S part:

$\to \bold{csc^4x-cot^4x}\\\\\to \bold{(csc^2x)^2-(cot^2x)^2}\\\\\to \bold{(csc^2x-cot^2x) (csc^2x+cot^2x)}\\\\\therefore \\\\(csc^2x-cot^2x =1)\\\\(csc^2x-1 = cot^2x)\\\\\to \bold{1 (csc^2x+cot^2x)}\\\\\to \bold{(csc^2x+csc^2x-1)}\\\\\to \bold{( 2csc^2x-1)}$

Hence proved that L.H.S= R.H.S.

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