Question

How to prove `(sintheta-costheta+1)/(sintheta+costheta-1)=sectheta+Tantheta` ?

Answer 1

We use `sin^2(x)+cos^2(x)=1` for the proof

Explanation:

Writing the Right-Hand side as
`1/cos(x)+sin(x)/cos(x)=(1+sin(x))/cos(x)`
and now by cross multiplication we get
`(sin(x)-cos(x)+1)cos(x)=sin(x)cos(x)-cos^2(x)+cos(x)`
and
`(sin(x)+cos(x)-1)(1+sin(x))=`
`sin(x)+cos(x)-1+sin^2(x)+sin(x)cos(x)-sin(x)`
Both sides are equal if
`sin^2(x)+cos^2(x)=1`

Answer 2

Please see below.

Explanation:

We take Left Hand Side :

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

Dividing (numerator and denominator) by `costheta`,

#LHS=((sintheta-costheta+1)/costheta)/((sintheta+costheta- 1)/costheta)#

`=(tantheta-1+sectheta)/(tantheta+1-sectheta`

#=(sectheta+tantheta-color(red)(1))/(1- sectheta+tantheta)...to[Put,color(red)(1=sec^2theta-tan^2theta)]#

#=(sectheta+tantheta-(color(red)(sec^2theta-tan^2theta)))/(1- sectheta+tantheta)#

#=((sectheta+tantheta)-(sectheta-tantheta) (sectheta+tantheta))/(1-sectheta+tantheta)#

#=((sectheta+tantheta)[1-(sectheta-tantheta)])/(1- sectheta+tantheta)#

#=((sectheta+tantheta)[1-sectheta+tantheta])/([1- sectheta+tantheta])#

`=sectheta+tantheta`

`=RHS`