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 $\theta&gt;90?$
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
$$
\cot(90^{\circ}-\theta)
=\frac{\cos(90^{\circ}-\theta)}{\sin(90^{\circ}-\theta)}
=\frac{\sin\theta}{\cos\theta}
=\tan\theta.
$$
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?