Question

How do you verify `tan^2(x) / (sec(x) - 1) = (sec(x) + 1)`?

Answer 1

The equation can be verified using difference of squares and some basic trigonometric definitions (`tan = sin/cos` and `sin^2+cos^2=1`)

Explanation:

`(sec(x)-1)*(sec(x)+1)`

`color(white)("XXXX")` `=sec^2(x)-1`

`color(white)("XXXX")` `=(sec(x)+1)(sec(x)-1)`

`color(white)("XXXX")` `=(1/(cos(x))+1)*(1/(cos(x))-1)`

`color(white)("XXXX")` `=(1+cos(x))/(cos(x))*(1-cos(x))/(cos(x)`

`color(white)("XXXX")` `=(1-cos^2(x))/(cos^2(x))`

`color(white)("XXXX")` `=(sin^2(x))/(cos^2(x))`

`color(white)("XXXX")` `= tan^2(x)`

Since `(sec(x)-1)*(sec(x)+1)= tan^2(x)`

`color(white)("XXXX")` `(tan^2(x))/(sec(x)-1) = sec(x)+1`

Note this proof relies on certain (obvious) values not being equal to zero (where there are divisions) but the operations hold true for values approaching zero and we can treat these cases as "removable".

Answer 2

The verification is done by Pythagorean Identity which is `tan^2(x)-sec^2(x)=1`

Explanation:

`Tan^2(x)/(secx-1)=(sec^2(x)-1)/(secx-1)`[from Pythagorean Identity ]

`" "color(white)(=)=(sec^2x-1^2)/(secx-1)`

`" "color(white)(=)=((secx+1)(secx-1))/(secx-1)`

[since `a^2-b^2=(a+b)(a-b)`]

`" "color(white)(=)=((secx+1)cancel((secx-1)))/(cancel((secx-1)))`

`" "color(white)(=)=secx+1`

Hence proved.