$\int \large\frac{1}{cosec^2 x. \sec ^2 x }$$dx$
By trignometry property
=> $\int \large\frac{cosec^2 x- \cot ^2 x}{cosec ^2 x \; sec ^2 x }$$dx$
=> $\int (\cos^2 x)- \large\frac{\cos ^2 x. \sin ^2 x . \cos ^2 x }{\sin ^2 x }.$$dx$
=> $\int (\cos ^ 2 x - \cos ^4 x)dx$
=> $\int \bigg(\large\frac{1+\cos 2 x}{2}\bigg)$$dx -\int \cos ^4 x$
=> $\int \bigg(\large\frac{1+\cos 2 x}{2}\bigg)$$dx -\int (1+\cos 2x)^2 dx$
=> $\int \bigg(\large\frac{1+\cos 2 x}{2}\bigg)$$dx -\large\frac{1}{4}$$ [(1+ \cos ^2 2x)+ 2 \cos 2x]$
=> $\int \large\frac{1}{2} +\frac{\cos 2 x}{2}-\frac{1}{4}-\frac{\cos ^2 2x}{4} -\frac{1}{2} $$\cos 2x $$dx -\int \bigg(\large\frac{1}{4} - \frac{\cos ^2 2x}{4} \bigg) $$dx$
=> $\int \large\frac{1}{4} $$(1- \cos^2 2x).dx$
=> $\int \large\frac{1}{4} $$\{1-(\frac{1}{2} (1+ \cos 4x))dx$
=> $\int \large\frac{1}{4} $$\bigg[1-\frac{1}{2}-\frac{\cos 4x}{2}\bigg]dx$
=> $\int \large\frac{1}{4} $$\{\large\frac{1}{4} \bigg\{\frac{1}{2} -\frac{\cos 4x}{2} \bigg] $$dx$
=> $\int \large\frac{1}{8} $$\bigg\{x-\large\frac{\sin 4x}{4}\bigg\} $$+c$
Hence a is the correct answer.