Draw the graph of `y=(sin2x)sqrt(1+tan^2x)` , find its domain and range.

Explanation

`y = (sin 2x) sqrt(1 + tan^(2)x)`
`= (2sin x cos x)|sec x|`
`={{:(2sin x"," " " cos x gt 0),(-2sin x"," " " cos x lt 0):}`
`={{:(2sin x"," " " x in 1^(st) and 4^(th) " quadrant"),(-2sin x"," " " x in 2^(nd) "and " 3^(rd) " quadrant" ):}`
Clearly, `f(pi + x) = f(x)`
So the period of the function is `pi`. First draw the graph for `x in ((-pi)/(2), (pi)/(2))`
`f((-pi)/(2)) = 2`
f(0) = 0
`f((pi)/(2)) = 2`
The graph of the function is shown in the following figure.

From the graph, the range of the function is (-2, 2).