To prove:
[tex]\frac{sin(x+\frac{\pi}{2})}{sin(\pi-x)}=cotx[/tex]
Note that:
sin(A + B) = sinAcosB + cosAsinB
sin(A - B) = sinAcosB - cosAsinB
Applying these rules to the given trigonometric expression
[tex]\frac{sin(x+\frac{\pi}{2})}{sin(\pi-x)}=\frac{sinxcos\frac{\pi}{2}+cosxsin\frac{\pi}{2}}{sin\pi cosx-cos\pi sinx}[/tex]
Note that:
cos(π) = -1
cos(π/2) = 0
sin(π) = 0
sin(π/2) = 1
Substitute these values into the expression above
[tex]\begin{gathered} \frac{s\imaginaryI n(x+\frac{\pi}{2})}{cos(\pi- x)}=\frac{s\imaginaryI nx(0)+cosx(1)}{(0)cosx-(-1)sinx} \\ \\ \frac{s\mathrm{i}n(x+\frac{\pi}{2})}{cos(\pi-x)}=\frac{cosx}{s\mathrm{i}nx} \\ \\ \frac{s\mathrm{i}n(x+\frac{\pi}{2})}{cos(\pi-x)}=cotx(Proved) \end{gathered}[/tex]
