Question 1133086

{{{cot^2(pi/2 -x)+1 }}}


start with {{{cot(pi/2 -x)}}} first:


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


we can use the formulas
 
{{{cos (A-B)=cos(A)cos(B)+sin(A)sin(B) }}}
and 
{{{sin(A-B)=sin(A)cos(B)-cos(A)sin(B)}}}


then you have


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


since {{{cos(pi/2)=0}}} and {{{sin(pi/2)=1}}} we have


{{{cot(pi/2 -x)=(0*cos(x)+1*sin(x) )/(1*cos(x)-0*sin(x))}}}


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


{{{cot(pi/2 -x)=tan(x)}}}


then {{{cot^2(pi/2 -x)=tan^2(x)}}}


and, then {{{cot^2(pi/2 -x)+1=tan^2(x)+1}}}