Question 196590
{{{tan(pi/2+x)=-cot(x)}}} Start with the given equation.



Note: I'm ONLY manipulating the left side.



{{{sin(pi/2+x)/cos(pi/2+x)=-cot(x)}}} Use the identity {{{tan(x)=sin(x)/cos(x)}}}



{{{(sin(pi/2)cos(x)+cos(pi/2)sin(x))/(cos(pi/2)cos(x)-sin(pi/2)sin(x))=-cot(x)}}} Expand


Note: {{{sin(A+B)=sin(A)cos(B)+cos(A)sin(B)}}} and {{{cos(A+B)=cos(A)cos(B)-sin(A)sin(B)}}}



{{{((1)cos(x)+(0)sin(x))/((0)cos(x)-(1)sin(x))=-cot(x)}}} Evaluate the sine of {{{pi/2}}} to get 1. Evaluate the cosine of {{{pi/2}}} to get 0



{{{(cos(x)+0)/(0-sin(x))=-cot(x)}}} Multiply



{{{(cos(x))/(-sin(x))=-cot(x)}}} Simplify



{{{-cot(x)=-cot(x)}}} Use the identity {{{cot(x)=1/tan(x)=cos(x)/sin(x)}}}



So this verifies the identity {{{tan(pi/2+x)=-cot(x)}}}