Question

How does one verify `sinx-cscx=-cos^2x/sinx`?

`sinx-cscx=-cos^2x/sinx`

Answer 1

See below.

Explanation:

It's best to leave one side alone, and simplify another side.

Let's leave the right alone, while simplifying the left.

Recall that `cscx=1/x:`

`sinx-1/sinx=-cos^2x/sinx`

Subtract the expressions on the left, taking `sinx` as the common denominator.

`(sin^2x-1)/sinx=-cos^2x/sinx`

Now, recall the identity

`sin^2x+cos^2x=1`

This identity also tells us that

`sin^2x-1=-cos^2x`. So, we can use this to simplify our numerator on the left:

`=cos^2x/sinx=-cos^2x/sinx`

So, the identity holds true.

Answer 2

See below

Explanation:

We want to prove that `sinx-cscx=-cos^2x"/"sinx`. We use the following identity

`cscx=1/sinx`

So

`sinx-cscx`

`=sinx-1/sinx`

`=(sin^2x-1)/sinx`

`=-(1-sin^2x)/sinx`

`=-cos^2x/sinx`, as required. `square`

Answer 3

See method below

Explanation:

`cscx=1/sinx` (definition)

We want to show that `sinx-1/sinx=-cos^2x/sinx`

Since both terms contain `sinx` we can factorize

`sinx-1/sinx = 1/sinx(sin^2x-1)`

Substitute 1 using the identity `1=cos^2x + sin^2x`

`1/sinx(sin^2x-(cos^2x + sin^2x))=1/sinx(-cos^2x)`

`sin^2x` cancels out.

`1/sinx(-cos^2x)=-cos^2x/sinx`

Therefore `sinx-1/sinx=-cos^2x/sinx`