Draw the graph of `f(x) = sec x + "cosec " x, x in (0, 2pi) - {pi//2, pi, 3pi//2}`
Also find the values of 'a' for which the equation `sec x + "cosec " x = a` has two distinct root and four distinct roots.

Explanation

We have `f(x) = sec x + "cosec " x, x in (0, 2pi) - {pi//2, pi, 3pi//2}`
Consider each of the intervals `(0, pi//2), (pi//2, pi), (pi, 3pi//2), (3pi//2, 2pi)`
For `x in (0, pi//2)`
When `x to 0^(+) or x to pi//2^(-), sec x + "cosec " x to oo`
Also in the `1^(st)` quadrant, sec x + cosec x gt 0
`f'(x) = sec x tan x - "cosec " x cot x`
`f'(pi//4) = 0,` so `x = pi//4` is clearly a point of minima.
`f(pi//4) = 2sqrt(2)`
For `x in (pi//2, pi)`
When `x to pi//2^(+), sec x + "cosec " x to - oo`
When `x to pi^(-), sec x + "cosec " x to oo`
Also `f'(x) + sec x tan x - "cosec " x cot x gt 0`
Hence f(x) is an increasing function.
`f(3pi//4) = 0`
For `x in (pi, 3pi//2)`
When `x to pi^(+), sec x + "cosec " x to -oo`
When `x to 3pi//2^(-), sec x + "cosec " x to -oo`
`f'(5pi//4) = 0, " so " x = 5pi//4` is clearly a point of maxima.
`f(5pi//4) = -2sqrt(2)`
For `x in (3pi//2, 2pi)`
When, `x to 3pi//2^(+), sec x + "cosec " x to oo`
When `x to 2pi^(-), sec x + "cosec " x to -oo`
Also `f'(x) + sec x tan x - "cosec " x cot x lt 0`
Hence f(x) is a decreasing function.
`f(7pi//4) = 0`
From the above discussion, the graph of the function is as shown in the following figure.