trigonometrical range...quite interesting

Let $\theta = x$

$\frac{ cosec^{2}x - cotx + 2cotx}{cosec^{2}x - cotx}$

$1 + \frac{2cotx}{cosec^2x - cotx}$

$1 + \frac{2}{ \frac{2}{sin2x} - 1}$ ( multiplying both numerator and denominator by tanx)

$1 + \frac{2sin2x}{2 - sin2x}$

$1 + \frac{2sin2x - 4 + 4}{2 - sin2x}$

$= \boxed{ \frac{4}{2 - sin2x} - 1}$

now since the range is represented by[a , b] , thus a is minimum value of it and b - maximum.

The faction is minimum when denominator is maximum , thus it is attained at $x = \frac{-\pi}{4}$ and the fraction is maximum when denominator is minimum which is attained at $x = \frac{\pi}{4}$

$[ \frac{1}{3} , 3]$

My method is bit different and involves calculus.

The given expression can be rewritten as $\dfrac{cot^{2}\theta +cot\theta +1}{cot^{2}\theta -cot\theta+1}$ Let $cot\theta$ = $x$ .

We see that Range of $cot\theta$ is $R$ so that x can take all real values.

So the expression becomes $\dfrac{x^{2} + x+1}{x^{2} -x+1}$ $=1+\dfrac{2x}{x^{2} -x+1}$

$Let f(x) = 1+\dfrac{2x}{x^{2} -x+1}$ $f'(x) = \dfrac{2(x^{2} -x+1)-(2x-1)2x}{(x^{2} -x+1)^{2}}$ At maxima or minima $f'(x) =0$ . Therefore $2(x^{2}-x+1) -(2x-1)2x=0$ $x=\pm 1$ When $x=1$ $f(x)=3$ When $x=-1$ $f(x)=\dfrac{1}{3}$ So $\boxed{\dfrac{b}{a} = 9}$

Given expression $=\frac{1+sin\theta cos\theta}{1-sin\theta cos\theta}$

$=\frac{2+sin2\theta }{2-sin2\theta}$

$=\frac{4}{2-sin2\theta}-1$ .

This expression is maximum when $sin2\theta = 1$ means $b=3$ and minimum when $sin2\theta = -1$ means $a=\frac{1}{3}$

So, $\frac{b}{a}=\boxed{9}$

I did it using quadratic. Converting cosec^2(@) into 1 + cot^2(@), then we can take cot(@) = x and solve for the range of the function.