PLZ HELP NOw

Please help me, why tan(90)=Infinity?

I typed this in calc: $tan(90)$

and it was equal to $infinity$.

Hi guest answerer, you have not answered, you have asked a new question.

You need to post a new question of your own.

you have asked a good question ... but you need to post it properly.

What is the residue modulo 13 of the sum of the modulo 13 inverses of the first 12 positive integers?
Express your answer as an integer from 0 to 12 , inclusive.

$\begin{array}{|rcll|} \hline 1\cdot { \color{red}1} &\equiv& 1 \pmod{13} \\ 2\cdot 2^{-1} \equiv \pmod{13} \equiv 2\cdot { \color{red}2^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 3\cdot 3^{-1} \equiv \pmod{13} \equiv 3\cdot { \color{red}3^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 4\cdot 4^{-1} \equiv \pmod{13} \equiv 4\cdot { \color{red}4^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 5\cdot 5^{-1} \equiv \pmod{13} \equiv 5\cdot { \color{red}5^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 6\cdot 6^{-1} \equiv \pmod{13} \equiv 6\cdot { \color{red}6^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 7\cdot 7^{-1} \equiv \pmod{13} \equiv 7\cdot { \color{red}7^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 8\cdot 8^{-1} \equiv \pmod{13} \equiv 8\cdot { \color{red}8^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 9\cdot 9^{-1} \equiv \pmod{13} \equiv 9\cdot { \color{red}9^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 10\cdot 10^{-1} \equiv \pmod{13} \equiv 10\cdot { \color{red}10^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 11\cdot 11^{-1} \equiv \pmod{13} \equiv 11\cdot { \color{red}11^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ 12\cdot 12^{-1} \equiv \pmod{13} \equiv 12\cdot { \color{red}12^{11}} \pmod{13} &\equiv& 1 \pmod{13} \\ \hline \end{array}$

$\begin{array}{|ll|} \hline & \text{sum}_{\text{of the modulo 13 inverses of the first 12 positive integers}} \\ =& 1^{11}+2^{11}+3^{11}+4^{11}+5^{11}+6^{11}+7^{11}+8^{11}+9^{11}+10^{11}+11^{11}+12^{11} \pmod{13} \\ =& 1+7+9+10+8+11+2+5+3+4+6+12 \pmod{13} \\ =& 1+2+3+4+5+6+7+8+9+10+11+12 \pmod{13} \\ =& \dfrac{1+12}{2}\cdot 12 \pmod{13} \\ =& 13 \cdot 6 \pmod{13} \\ \mathbf{=}& \mathbf{0 \pmod{13}} \\ \hline \end{array}$