Question

How do you prove `tan^-1x+tan^-1(1/x)=pi/2` for x>0?

Answer 1

we first need to know the following relationships between `tantheta` &`cot theta`

we have:

`cottheta=1/tantheta`

also

`cottheta=tan(pi/2-theta)`

Explanation:

let `y=tan^(-1)x=>x=tany`

`x=tany=>1/x=1/tany=coty`

`1/x=coty=tan(pi/2-y)`

`:.pi/2-y=tan^(-1)(1/x)`

but `y=tan^(-1)x`

so `pi/2=tan^(-1)(1/x)+tan^(-1)x`

as reqd.

Answer 2

Take the addition formula for tangent:

`tan(a+b)=(tan(a)+tan(b))/(1-tan(a)tan(b))`

Rearranging:

`a+b=tan^-1((tan(a)+tan(b))/(1-tan(a)tan(b)))`

Let `m=tan(a)` and `n=tan(b)`. These both imply that `a=tan^-1(m)` and `b=tan^-1(n)`.

Plugging in all these values into the before formula gives:

`tan^-1(m)+tan^-1(n)=tan^-1((m+n)/(1-mn))`

This is applicable to the question, which asks about `tan^-1(x)+tan^-1(1/x)`. According to the rule we've just derived:

`tan^-1(x)+tan^-1(1/x)=tan^-1((x+1/x)/(1-x(1/x)))`

`tan^-1(x)+tan^-1(1/x)=tan^-1((x+1/x)/(1-1))`

Since the denominator inside the inverse tangent function on the right-hand side is `0`, we see that the argument of the function is undefined.

For `x>0` and using the restricting the range of `tan^-1(x)` to `(-pi/2,pi/2)`, we see that tangent is undefined at `pi/2`, so the inverse tangent of an undefined value is also `pi/2`.

Therefore:

`tan^-1(x)+tan^-1(1/x)=pi/2`