# y=4+cotx-2cscx: Find the equation for the tangent at x=pi/2y=4+cotx-2cscx, x = pi/2 Thank you very much in advanced :DD

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To find the equation of the tangent line, determine the point of tangencey and its slope.

To do so, susbtitute `x=pi/2` to the equation of the curve.

`y=4+cotx-2cscx=4+cot(pi/2)-2csc(pi/2)`

Note that `cot(pi/2)=0` and `csc(pi/2)=1` .

`y=4+0-2(1)=2`

Hence, the point of tangency is `(pi/2,2)` .

The slope of the tangent line is the same as the slope of the curve at the point of tangency.

To determine the slope of the curve take the derivative of y.

`y=4+cotx-2cscx`

`y'=(4+cotx-2cscx)'`

Note that `(c)'=0` (where c is a constant) , `(cotu)=-csc^2u*u'` and `cscu=-cscucotu*u'` .

`y'=0-csc^2x*1 -2(-cscxcotx*1)=-csc^2x+2cscxcotx`

Subsitute `x=pi/2` .

`y'=-csc^2(pi/2)+2csc(pi/2)cot(pi/2)=-(1)^2+2*1*0`

`y'=-1`

Hence, the slope of the tangent line is -1.

Now that the slope of the line and point of tangency is known, apply the point-slope formula of a line.

`y-y_1=m(x-x_1)`

where m is the slope and `(x_1, y_1)` is the given point.

Substitute the slope m=-1 and the point `(pi/2,2)` .

`y-2=-1(x-pi/2)`

`y-2=-x+pi/2`

`y=-x+pi/2+2`

`y=-x+pi/2+4/2`

`y=-x+(4+pi)/2`

**Hence, the equation of the tangent line is `y=-x+(4+pi)/2` .**