Answer: \(a=\dfrac{\pi}{6}\).

Use

\(\operatorname{cosec}\theta=\frac1{\sin\theta},\quad \operatorname{cot}\theta=\frac{\cos\theta}{\sin\theta},\quad \tan\theta=\frac{\sin\theta}{\cos\theta}.\)

Then

\(\operatorname{cot}\theta+\tan\theta = \frac{\cos\theta}{\sin\theta} + \frac{\sin\theta}{\cos\theta} = \frac{\cos^2\theta+\sin^2\theta}{\sin\theta\cos\theta} = \frac1{\sin\theta\cos\theta}.\)

Therefore

\(\frac{\operatorname{cosec}\theta}{\operatorname{cot}\theta+\tan\theta} = \frac{1/\sin\theta}{1/(\sin\theta\cos\theta)} = \cos\theta.\)

Replacing \(\theta\) by \(2\theta\),

\(\frac{\operatorname{cosec}2\theta}{\cot2\theta+\tan2\theta}=\cos2\theta.\)

Hence

\(\int_0^a \cos2\theta\,d\theta=\frac{\sqrt3}{4}.\)

So

\(\left[\frac12\sin2\theta\right]_0^a=\frac{\sqrt3}{4}.\)

Thus

\(\frac12\sin2a=\frac{\sqrt3}{4}, \qquad \sin2a=\frac{\sqrt3}{2}.\)

Since \(0\lt a\lt \dfrac{\pi}{4}\), we have \(0\lt 2a\lt \dfrac{\pi}{2}\). Therefore

\(2a=\frac{\pi}{3},\)

and

\(a=\frac{\pi}{6}.\)