Prove : ` ((1+cot^2 A)/(1+tan^2 A​))=((1−cotA)/(1−tanA​))^2=tan^2 A`

Explanation

From the graph , ` f' lt 0" for " x in (1, 3) cup (5, 6)` where f decreases.
From the graph,` f' gt 0 " for " x in (0, 1) in (3, 5)` where f increases.
At x = 1 and x = 5, f' changes sign from '+' to '-', so it is the point of local maxima.
At x = 3, f'changes sign from '-' to '+' , so it is the point of local minima.
Slope of tangent to curve f' is '-' for ` x in (0, 2), i.e. f''lt 0`.
Slope of tangent to curve f ' is '+' for ` x in (2, 4), i.e. f'' gt 0`.
That is , f'' changes sign at x = 2 with f''(2) = 0, so x = 2 is the point of inflection.
Further slope of the tangent to the curve f' is '-' for ` x in (4, 6) , i.e. f'' lt 0`.
That is , f'' changes sign at x = 4 again f ''(4) = 0, so x = 4 is the point of inflection.