Question

Question #dd613

Answer 1

`sinx`

Explanation:

First simplify the inverse trig identities.
`cscx=1/sinx`; `cotx=cosx/sinx`
Plugging those in you get
`1/sinx-cosx(cosx/sinx)` multiply
`1/sinx-cos^2x/sinx` combine them together
`(1-cos^2x)/sinx` Use trig identities of squared trig functions
`sin^2x+cos^2x=1 => sin^2x=1-cos^2x`
`sin^2x/sinx=>sinx` Reduce

Answer 2

`cscx-cosx*cotx=sinx`

Explanation:

`cscx-cosx*cotx`

=`1/sinx-cosx*cosx/sinx`

=`[1-(cosx)^2]/sinx`

=`(sinx)^2/sinx`

=`sinx`