Answer:True.Step-by-step explanation:Let's use the picture I made.I used degrees instead...tan(90-x)= b/a . I did opposite over adjacent for the angle labeled 90-x which is that angle's measurement. cot(x)=b/a . I did adjacent over opposite for the angle labeled 90 which is that angle's measurement. Now this is also known as a co-function identity. [tex]\tan(\frac{\pi}{2}-x)[/tex]Rewrite using quotient identity for tangent[tex]\frac{\sin(\frac{\pi}{2}-x)}{\cos(\frac{\pi}{2}-x)}[/tex]Rewrite using difference identities for sine and cosine[tex]\frac{\sin(\frac{\pi}{2})\cos(x)-\sin(x)\cos(\frac{\pi}{2})}{\cos(\frac{\pi}{2})\cos(x)+\sin(\frac{\pi}{2})\sin(x)}[/tex]sin(pi/2)=1 while cos(pi/2)=0[tex]\frac{1 \cdot \cos(x)-\sin(x) \cdot 0}{0 \cdot \cos(x)+1 \cdot \sin(x)}[/tex]Do a little basic algebra[tex]\frac{\cos(x)-0}{0+\sin(x)}[/tex]More simplification[tex]\frac{\cos(x)}{\sin(x)}[/tex]This is quotient identity for cotangent [tex]\cot(x)[/tex]