Question

How do you simplify `Tan(sec^-1(u))`?

Answer 1

`y/x`

Explanation:

Trigonometric functions of an angle can be thought of as ratios of the sides of a triangle and its inverse functions are functions of a ratio of that triangle that return an angle.

Let there be a right triangle with vertical leg `y`, horizontal leg `x` and hypotenuse `r`.
Let an angle `theta` be defined as the angle formed from the intersection of `x` and `r` measured counter-clockwise.

`sec(theta) = 1/cos(theta) = r/x`
so `sec^-1(u)=sec^-1(r/x)=theta`
The problem now becomes `tan(theta)`.

We know that `tan(theta)=sin(theta)/cos(theta)=(y/cancel(r))/(x/cancel(r))` = `y/x`

Answer 2

see below

Explanation:

Let `sec^-1(u)` be equal to some `theta`.
Thus,
`sec theta=u`
Squaring both sides,we get,
`sec^2 theta=u^2`
`tan^2theta+1`=`u^2` [we know,`tan^2theta+1=sec^2 theta`]
Thus,
`tan theta=+-sqrt(u^2-1)`

Getting back to the original question,we have,
`tan(sec^-1(u))` which on simplification gives,
`tan theta` which is equal to `+-sqrt(u^2-1)`