Question

Using derivative, prove that: sec^–1x + cosec^–1x = `pi/(2)`    ...[for |x| ≥ 1]

Answer 1

Let f(x) = sec^–1x + cosec^–1x  for |x| ≥ 1  ...(1)
Differentiating w.r.t. x, we get
f'(x) = `"d"/"dx"(sec^-1 x + "cosec"^-1 x)`

= `"d"/"dx"(sec^-1 x) + "d"/"dx"("cosec"^-1 x)`

= `(1)/(xsqrt(x^2 - 1)) - (1)/(xsqrt(x^2 - 1)`
= 0
Since, f'(x) = 0, f(x) is a constant function.
Let f(x) = k.
For any value of x,f(x) = k, where |x| > 1
Let x = 2.
Then, f(2) = k                                      ...(2)
From (1),f(2) = sec^–1(2) + cosec^–1(2)
= `pi/(3) + pi/(6) = pi/(2)`
∴ k = `pi/(2)`                                  ...[By (2)]
∴ f(x) = k = `pi/(2)`
Hence, sec^–1x + cosec^–1x = `pi/(2)`.      ...[By (1)]