Proving that $\tan \theta=\cot(90^\circ-\theta)$ when $\theta>90^\circ$ 
I'm asked to prove $\tan \theta=\cot (90-\theta)$


So $\tan \theta =\frac{a}{b}=\cot(90^\circ-\theta)$
But what if $\theta>90?$
 A: Use the identities
$$
\sin(x-y)=\sin x\cos y-\cos x\sin y
$$
and 
$$
\cos(x-y)=\cos x\cos y+\sin x\sin y
$$
to deduce that
$$
\sin(90^{\circ}-\theta)=\cos\theta;\quad \cos(90^{\circ}-\theta)=\sin\theta.
$$
Hence
$$
\cot(90^{\circ}-\theta)
=\frac{\cos(90^{\circ}-\theta)}{\sin(90^{\circ}-\theta)}
=\frac{\sin\theta}{\cos\theta}
=\tan\theta.
$$
A: Let $0^\circ<\alpha<90^\circ$.
Second and third quadrants:
If $\theta=180^\circ\pm\alpha$, then
$$\begin{align}
\text{RHS}&=\cot(90^\circ-\theta)\\
&=\cot(90^\circ-(180^\circ\pm\alpha))\\
&=\cot(-90^\circ\mp\alpha)\\
&=\cot(-(90^\circ\pm\alpha))\\
&=-\cot(90^\circ\pm\alpha)\\
&=\mp\tan\alpha\\
\text{LHS}&=\tan\theta\\
&=\tan(180^\circ\pm\alpha)\\
&=\mp\tan\alpha\\
&=\text{RHS}
\end{align}$$
Fourth quadrant:
If $\theta=-\alpha$, then
$$\begin{align}
\text{RHS}&=\cot(90^\circ-\theta)\\
&=\cot(90^\circ+\alpha)\\
&=\cot(180^\circ-(90^\circ -\alpha))\\
&=-\cot(90^\circ-\alpha))\\
&=-\tan\alpha\\
\text{LHS}&=\tan\theta\\
&=\tan(-\alpha)\\
&=-\tan\alpha\\
&=\text{RHS}
\end{align}$$
A: You can prove this using the fact $\sin(\theta) = \cos(90^\circ - \theta)$ for all $\theta \in \Bbb R$:
$$ \begin{align*}
\tan(90^\circ-  \theta) &= \frac{\sin(90^\circ - \theta)}{\cos(90^\circ -
 \theta)} \\
&= \frac{\cos(\theta)}{\sin( \theta)} \\
&= \cot(\theta)
\end{align*}. $$
A: I think it should be noted that the $\tan(\theta) = \frac{opp}{adj}$ and $\cot(\theta) = \frac{adj}{opp}$ are only defined for acute triangles. If $\theta>90$, then to extend the notion of trigonometric functions to obtuse angles, let $r$ be the distance $|OP|$ between the origin $O$ and $P$ is the point at $(x,y)$. Then, $\tan(\theta) = \frac{y}{x}$ and $\cot(\theta) = \frac{x}{y}$. This line of reasoning should be suggestive, since you are considering an angle larger than 90 degrees.
A: Have a look at this figure from the Wikipedia article on trigonometric functions. Tangent is the directed length of the segment of the circle's tangent line (tangent at the correct angle) between the circle and the $x$-axis. Cotangent is the length of the segment of the tangent line that connects to the $y$-axis. The direction, and thus the sign, of each is defined such that it alternates as you go from quadrant to quadrant.
A: At a point the angle of elevation of the top of a tower is such that its tangent is 5/12 . On walking 80m towards the tower, the cotangent of the angle of elevation of the top of the tower is 4/3. The height of the tower will be how much m?
