By considering a right-angled triangle, Prove that if $A$ is an acute angle, then $\tan(90-A)=\frac1{\tan A }$.

Question

By considering a right-angled triangle, Prove that if $A$ is an acute angle, then $\tan(90-A)=\dfrac1{\tan A}$.

I'm not too sure with this question, would it be a good starting point by using the identity $\tan x=\dfrac{\sin x}{\cos x}$ ?

Answer 1

$\tan(90-A) = \sin(90°-A)/\cos(90°-A) = \cos(A) / \sin(A) = 1/\tan(A)$

Answer 2

If the legs are $a$ and $b$ opposite corresponding angles $A$ and $B$ then $\tan(A) = a/b$ and $\tan(B) = b/a$.

But $A+B = 90$ so $\tan(B) =\tan(90-A) =b/a =1/(a/b) =1/\tan(A) $.