Question

How to prove? A)√((1-sin⁡θ)/(1+sin⁡θ ))=secθ-tanθ B) sinA-sinB/cosB-coaA = cot(A+B/2)

Answer 1

See the proof below

Explanation:

`"(A)"`

`LHS=sqrt((1-sintheta)/(1+sintheta))`

`=sqrt(((1-sintheta)(1-sintheta))/((1+sintheta)(1-sintheta)))`

`=sqrt(((1-sintheta)^2)/(1-sin^2theta))`

`=sqrt(((1-sintheta)^2)/(cos^2theta))`

`=(1-sintheta)/costheta`

`=1/costheta-sintheta/costheta`

`=sectheta-tantheta`

`=RHS`

`QED`

`"(B)"`

`LHS=(sinA-sinB)/(cosB-cosA)`

`=(2cos((A+B)/2)sin((A-B)/2))/(-2sin((A+B)/2)sin(B-A)/2)`

`=cos((A+B)/2)/(sin((A+B)/2`

`=cot((A+B)/2)`

`=RHS`

`QED`

Answer 2

Formulae to use:
`(1)1-sin^2theta=cos^2theta`
`(2)1/costheta=sectheta,..(3)sintheta/costheta=tantheta`
`(4)sinx-siny=2cos((x+y)/2)sin((x-y)/2)`
`(5)cosx-cosy=-2sin((x+y)/2)cos((x-y)/2)`
Pls. see below

Explanation:

`color(red)([A]sqrt((1-sintheta)/(1+sintheta))=sectheta-tantheta`

`LHS=sqrt(((1-sintheta)/(1+sintheta))xx((1-sintheta)/(1-sintheta))`

#=sqrt((1-sintheta)^2/(1-sin^2theta))=sqrt((1- sintheta)^2/(cos^2theta))...toapply (1)#

`=(1-sintheta)/costheta`

`=1/costheta-sintheta/costheta`

`=sectheta-tantheta....to Apply(2)and(3)`

`=RHS`

`color(red)([B] (sinA-sinB)/(cosB-cosA)=cot((A+B)/2)`

#LHS=(cancel2cos((A+B)/2)sin((A-B)/2))/(- cancel2sin((B+A)/2)sin((B-A)/2))..to Apply(4)and(5)#

#=(cancel2cos((A+B)/2)cancelsin((A- B)/2))/(cancel2sin((B+A)/2)cancelsin((A-B)/2)#

`=cos((A+B)/2)/sin((A+B)/2).... to` But,.`cosx/sinx=cotx`

`=cot((A+B)/2).`

`=RHS`