Respuesta :
[tex]\bf \textit{Pythagorean Identities}
\\\\
sin^2(\theta)+cos^2(\theta)=1
\\\\
1+cot^2(\theta)=csc^2(\theta)
\\\\
1+tan^2(\theta)=sec^2(\theta)\\\\
-------------------------------[/tex]
[tex]\bf 1+cot^2(x)=csc^2(x)\implies cot^2(x)-csc^2(x)=-1 \\\\\\ \boxed{csc^2(x)-cot^2(x)=1}\qquad \qquad A~\otimes\\\\ -------------------------------\\\\ 1+tan^2(x)=sec^2(x)\implies \boxed{tan^2(x)=sec^2(x)-1}\qquad \qquad B~\checkmark \\\\ -------------------------------\\\\ sin^2(x)+cos^2(x)=1\implies \boxed{sin^2(x)=1-cos^2(x)}\qquad \qquad C~\checkmark \\\\ -------------------------------\\\\ \boxed{sin^2(x)+cos^2(x)=1}\qquad \qquad D~\otimes[/tex]
[tex]\bf 1+cot^2(x)=csc^2(x)\implies cot^2(x)-csc^2(x)=-1 \\\\\\ \boxed{csc^2(x)-cot^2(x)=1}\qquad \qquad A~\otimes\\\\ -------------------------------\\\\ 1+tan^2(x)=sec^2(x)\implies \boxed{tan^2(x)=sec^2(x)-1}\qquad \qquad B~\checkmark \\\\ -------------------------------\\\\ sin^2(x)+cos^2(x)=1\implies \boxed{sin^2(x)=1-cos^2(x)}\qquad \qquad C~\checkmark \\\\ -------------------------------\\\\ \boxed{sin^2(x)+cos^2(x)=1}\qquad \qquad D~\otimes[/tex]