Three Trigonometry Functions.

The answer is very close to ( $1$ ) but it isn't ( $1$ ). If you got tricked by your calculator, see bottom.

Now, ( $1°$ ) is a really small angle. Visualize the triangle corresponding to ( $sin(1°)$ ). The hypotenuse is almost equal to the adjacent, making ( $sin(1°)$ ) almost, but is $\textbf{not}$ , equal to ( $0$ ).

Then, ( $tan(sin(1°))$ ). You can use the above method to find that $tan(\textbf{near zero})$ is also near, but $\textbf{not}$ ( $0$ ).

Similarly, you can do the same: $cos(\textbf{near zero})$ and find that it equals very closely to ( $1$ ) but it $\textbf{isn't}$ ( $1$ ).

The very fancy looking summation thing is, when you work it out, just a very very close approximation of ( $1$ ). You can easily tell that it is not the answer as it can be expressed in a closed form.

What your calculator is doing when it shows ( $1$ ) is that, since $tan(sin(1°))$ is so close to ( $0$ ), your calculator just assumes that $cos(tan(sin(1°)))=1$