Answer
$$\dfrac{1}{12}$$
Work Step by Step
The Taylor series can be written as follows: $\sin x= x-\dfrac{x^3}{3!}+\dfrac{ x^5}{5!}-...........$ and $\cos x=1-\dfrac{x^2}{2}+\dfrac{ x^3}{3}-....$
$$\lim\limits_{t \to 0} \dfrac{1}{2-2 \cos t}-\dfrac{1}{t^2}=\lim\limits_{t \to 0} \dfrac{t^2-2+2 \cos t }{2t^2 (1-\cos t)} \\=\lim\limits_{t \to 0} \dfrac{t^2-2+2 (1-\dfrac{t^2}{2!}+\dfrac{ t^4}{4}-....) }{2t^2 (1-1+\dfrac{t^2}{2!}+\dfrac{ t^4}{4}-....)} \\=\dfrac{1}{12}$$
