Question

CscAsecA-cotA??

Answer 1

This simplifies to `tanA`.

Explanation:

We start by rewriting in sine and cosine:

`= 1/sinA(1/cosA) - cosA/sinA`

`= 1/(sinAcosA) - cosA/sinA`

`= 1/(sinAcosA) - cos^2A/(sinAcosA)`

`= (1 -cos^2A)/(sinAcosA)`

`=sin^2A/(sinAcosA)`

`= sinA/cosA`

`=tanA`

Hopefully this helps!

Answer 2

`tanA`

Explanation:

`cscAsecA-cotA`

`1/sinA*1/cosA-cosA/sinA`

`1/(sinAcosA)-cosA/sinA`

`1/(sinAcosA)-cosA/sinA(cosA/cosA)`

`1/(sinAcosA)-cos^2A/(sinAcosA)`

`(1-cos^2A)/(sinAcosA)`

From the pythagorean identities, we know that `1-cos^2A = sin^2A`.
`sin^2A/(sinAcosA)`

`sinA/cosA`

`tanA`

Hope this helps!