Question

Question #cab60

Answer 1

Recall the following identities:

`1. color(red)(secx=1/cosx)`

`2. color(darkorange)(tanx=sinx/cosx)`

`3. color(blue)(cotx=cosx/sinx)`

`4. color(purple)(sin^2x+cos^2x=1)`

Given the following identity, start the proof by working on the left side.

`secx/cosx-tanx/cotx=1`

Left side:

`(color(red)(1/cosx))/cosx-(color(darkorange)(sinx/cosx))/(color(blue)(cosx/sinx))`

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

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

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

`=cos^2x/cos^2x`

`=color(green)(|bar(ul(color(white)(a/a)1color(white)(a/a)|)))`

`:.`, left side`=`right side.

Answer 2

`secx/cosx -tan x/cot x`

`=sec^2 x - tan^2 x`

`=1`

Explanation:

LHS: `secx/cosx -tan x/cot x`

`=secx xx 1/cosx - tan x xx 1/cot x`

`=secx xx secx - tan x xx tan x`

`=sec^2 x - tan^2 x`

`=1`