Question

Question #ca9fe

Answer 1

`LHS=cscA*sin2A-secA`

`=cscA*(2sinA*cosA)-secA`

`=1/sinA*(2sinAcosA)-1/cosA`

`=2cosA-1/cosA`

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

`=(cos2A)/cosA`

`=cos2A*secA=RHS`

Answer 2

Use trigonometric identities as done below

Explanation:

We will make use of the identities below, found here and here . As for why we choose these identities, I can only refer to practice and intuition.
1. `csc(x) = 1/sin(x)`
2. `sec(x) = 1/cos(x)`
3. `sin(2x) = 2sin(x)cos(x)`
4. `cos(2x) = cos^2(x) - sin^2(x)`
5. `sin^2(x) + cos^2(x) = 1`

Rewrite `csc(A)sin(2A)-sec(A)` using identies 1. and 2.
`csc(A)sin(2A)-sec(A) =`
`sin(2A)/sin(A)-1/cos(A)`
use identity 3.
`(2sin(A)cos(A))/sin(A)-1/cos(A)`
`2cos(A) - 1/cos(A)`
Put on common denominator by multiplying the first term by `cos(A)`
`(2cos^2(A) - 1)/cos(A)`
Substitute the `1`, using identity 5.
`(2cos^2(A) - (cos^2(A) + sin^2(A)))/cos(A)`
Simplify (and mind the sign when removing the parenthesis)
`(cos^2(A) - sin^2(A))/cos(A)`
Use identity 4 and write the denominator as a factor, for clarity
`cos(2A)1/cos(A)`
Use identity 2.

`cos(2A)sec(A)`

and we have arrived at what we wanted to show.