Question

Using derivative, prove that: tan ^–1x + cot^–1x = `pi/(2)`

Answer 1

Let f(x) = tan ^–1x + cot ^–1x                    ...(1)
Differentiating w.r.t. x, we get
f'(x) = `"d"/"dx"(tan^-1 x + cot^-1 x)`

= `"d"/"dx"(tan^-1 x) + "d"/"dx"(cot^-1 x)`

= `(1)/(1 + x^2) - (1)/(1 + x^2)`
= 0
Since f'(x) = 0, f(x) is a constant function.
Let f(x) = k.
For any value of x, f(x) = k
Let x = 0.
Then f(0) = k                                       ...(2)
From (1), f(0) = tan^–1(0) + cot^–1(0)
= `0 + pi/(2) = pi/(2)`
∴ k = `pi/(2)`                                   ...[By (2)]
∴ f(x) = k = `pi/(2)`
Hence, tan^–1x + cot^–1x = `pi/(2)`.   ...[By (1)]