Question

Please help to simplify as far as possible? `(cosA)/(1-sinA) -tan A`

Answer 1

`1/cos(x)`

Explanation:

`tan(x)=sin(x)/cos(x)`
`sin^2(x)+cos^2(x)=1`
`cos^2(x)/(cos(x)(1-sin(x)))-sin(x)(1-sin(x))/(cos(x)(1-sin(x)))`
`=(cos^2(x)-sin(x)+sin^2(x))/(cos(x)(1-sin(x)))`
`=cancel(1-sin(x))/(cos(x)(cancel(1-sin(x))))`
`=1/cos(x)`

Answer 2

`secA`

Explanation:

#cosA/(1-sinA)-tanA=cosA/(1-sinA)-sinA/cosA =(cos^2A-sinA+sin^2A)/((1-sinA)cosA) =((color(red)(cos^2A+sin^2A)-sinA)/((1-sinA)cosA) =cancel(color(red)1-sinA)/(cancel((1-sinA))cosA)=1/cosA=secA#
Hint: `color(red)(cos^2A+sin^2A=1)`

Answer 3

`secA`

Explanation:

`"using the "color(blue)"trigonometric identities"`

`•color(white)(x)sin^2A+cos^2A=1`

`•color(white)(x)tanA=sinA/cosA" and "secA=1/cosA`

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

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

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

`=(cancel((1-sinA)))/(cosAcancel((1-sinA))`

`=1/cosA=secA`