If the chord $y=mx+1$ of the circle $x^2+y^2=1$ subtends  an angle of measure ${45}^0$ at the major segment of the circle then $m=$

Chord makes ${90}^0$ at the centre $(0,0)$ 

Homogenise the circle equation with the line $y-mx=1$ and then sum of the coefficients of $x^2$ and $y^2$ must be equal to zero

It implies that 

$\begin{array}{rcl}{x}^2+y^2& =& {\left(y-mx\right)}^2\\ x^2+y^2& =& m^2{x}^2+y^2-2mxy\\ x^2\left(1-m^2\right)+2mxy& =& 0\end{array}$

The above equaiton represent pair of lines passing through the origin and the end points of the chord

Since the angle between the above two lines is right angles

$\begin{array}{rcl}1-m^2+0& =& 0\\ m^2& =& 1\\ m& =& \pm 1\end{array}$