# How do you find at equation of the tangent line to the graph #y=4+cotx-2cscx# at x=pi/2?

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To find the equation of the tangent line to the graph y=4+cotx-2cscx at x=pi/2, we need to find the derivative of the function and evaluate it at x=pi/2. The derivative of y=4+cotx-2cscx is dy/dx = -csc^2(x) + csc(x)cot(x). Evaluating this at x=pi/2, we get dy/dx = -1 + 0 = -1.

The slope of the tangent line is equal to the derivative at the given point, which is -1. To find the equation of the tangent line, we use the point-slope form: y - y1 = m(x - x1), where (x1, y1) is the given point (pi/2, 4+cot(pi/2)-2csc(pi/2)) and m is the slope (-1).

Substituting the values, we have y - (4 + cot(pi/2) - 2csc(pi/2)) = -1(x - pi/2). Simplifying further, we get y - (4 + 0 - 2) = -1(x - pi/2).

Simplifying the equation, we have y - 2 = -x + pi/2. Rearranging, we get y = -x + pi/2 + 2.

Therefore, the equation of the tangent line to the graph y=4+cotx-2cscx at x=pi/2 is y = -x + (pi/2 + 2).

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