Step 1 :Given the equation of the circle is \(x^{2}+y^{2}=29\) and the point is \((-2,-5)\).
Step 2 :Rearrange the equation of the circle to get \(y^{2}=29-x^{2}\).
Step 3 :Take the square root of both sides to get \(y=\sqrt{29-x^{2}}\).
Step 4 :Find the derivative of y with respect to x to get \(y'=-\frac{x}{\sqrt{29-x^{2}}}\).
Step 5 :Substitute \(x=-2\) into the derivative to find the slope of the tangent line at the point \((-2,-5)\), which is \(y'(-2)=\frac{2}{5}\).
Step 6 :Substitute the point \((-2,-5)\) and the slope \(\frac{2}{5}\) into the equation of a line \(y=mx+b\) to find the y-intercept, which is \(b=-4.2\).
Step 7 :\(\boxed{The equation of the tangent line at the point (-2,-5) is y=\frac{2}{5}x-4.2}\).