Question

Question #e026e

Answer 1

See the proof below

Explanation:

We need

`cscx=1/sinx`

`cotx=cosx/sinx`

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

Therefore

`LHS =(1/(cscx-cotx))-(1/(cscx+cotx))`

`=(1/(1/sinx-cosx/sinx))-(1/(1/sinx+cosx/sinx))`

`=(sinx/(1-cosx))-(sinx/(1+cosx))`

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

`=sinx/sin^2x*2cosx`

`=2/(sinx/cosx)`

`=2/tanx`

`=RHS`

`QED`

Answer 2

`(1/(cscx - cotx)) -(1/(cscx + cotx)) = 2/tanx`

Multiply the fractions on the left side by the conjugates of the denominators (which is the same as multiplying by one):

`"Left side:"`
`frac{1}{cscx-cotx} color(blue)(*frac{cscx+cotx}{cscx+cotx}) - frac{1}{cscx + cotx} color(blue)(*frac{cscx-cotx}{cscx-cotx})`

`"LS"= frac{(cscx+cotx) - (cscx - cotx)}{csc^2 x -cot^2 x}`

Using the pythagorean trig identity, we know that `csc^2x - cot^2x = 1`
`"LS" = frac{color(red)(cscx) + cotx color(red)(-cscx) + cotx}{color(blue)((csc^2x - cot^2x))}`

`"LS" = 2cotx`

`"LS" = 2/tanx`

`="Right side"`

`QED`