Question

Discuss the following derivation:

$\tan \left(\theta +\frac{\mathrm{\pi }}{2}\right)=\frac{\tan \left(\theta \right)+\tan \left(\frac{\mathrm{\pi }}{2}\right)}{1-\tan \left(\theta \right)\tan \left(\frac{\mathrm{\pi }}{2}\right)}=\frac{{\displaystyle \frac{\tan \left(\theta \right)}{\tan \left(\frac{\mathrm{\pi }}{2}\right)}}+1}{{\displaystyle \frac{1}{\tan \left(\frac{\mathrm{\pi }}{2}\right)}}-\tan \left(\theta \right)}=\frac{0+1}{0-\tan \left(\theta \right)}=\frac{1}{-\tan \left(\theta \right)}=-cot\left(\theta \right)$

Short answer

The derivation is not correct becouse it consists of $\tan \frac{\mathrm{\pi }}{2}$, which is undefined and makes the further steps undefined

The correct derivation is

$$\begin{gathered} \tan \left(\theta +\frac{\mathrm{\pi }}{2}\right)=\frac{\sin \left(\theta +\frac{\mathrm{\pi }}{2}\right)}{\cos \left(\theta +\frac{\mathrm{\pi }}{2}\right)} \\ =\frac{\sin \theta \cos \frac{\mathrm{\pi }}{2}+\cos \theta \sin \frac{\mathrm{\pi }}{2}}{\cos \theta \cos \frac{\mathrm{\pi }}{2}-\sin \theta \sin \frac{\mathrm{\pi }}{2}} \\ =\frac{0+\cos \theta }{0-\sin \theta } \\ =cot\theta \end{gathered}$$

Step-by-step solution

Step 1. Given data

The given derivation is

$\tan \left(\theta +\frac{\mathrm{\pi }}{2}\right)=\frac{\tan \left(\theta \right)+\tan \left(\frac{\mathrm{\pi }}{2}\right)}{1-\tan \left(\theta \right)\tan \left(\frac{\mathrm{\pi }}{2}\right)}=\frac{{\displaystyle \frac{\tan \left(\theta \right)}{\tan \left(\frac{\mathrm{\pi }}{2}\right)}+1}}{{\displaystyle \frac{1}{\tan \left(\frac{\mathrm{\pi }}{2}\right)}-\tan \left(\theta \right)}}=\frac{0+1}{0-\tan \left(\theta \right)}=\frac{1}{-\tan \left(\theta \right)}=-cot\left(\theta \right)$

Step 2. justification of step

Consider the step

$\tan \left(\theta +\frac{\mathrm{\pi }}{2}\right)=\frac{\tan \left(\theta \right)+\tan \left(\frac{\mathrm{\pi }}{2}\right)}{1-\tan \left(\theta \right)\tan \left(\frac{\mathrm{\pi }}{2}\right)}$

In this step, a sum formula for the tangent function is usedlocalid="1646503250220" $\tan \left(\alpha +\beta \right)=\frac{\tan \left(\alpha \right)+\tan \left(\beta \right)}{1-\tan \left(\alpha \right)\tan \left(\beta \right)}$

Where$$\begin{gathered} \alpha =\theta \\ \beta =\frac{\mathrm{\pi }}{2} \end{gathered}$$

So this step is justified

Step 3. Justification of step

Consider the step

$\frac{\tan \left(\theta \right)+\tan \left(\frac{\mathrm{\pi }}{2}\right)}{1-\tan \left(\theta \right)\tan \left(\frac{\mathrm{\pi }}{2}\right)}=\frac{{\displaystyle \frac{\tan \left(\theta \right)}{\tan \left(\frac{\mathrm{\pi }}{2}\right)}+1}}{{\displaystyle \frac{1}{\tan \left(\frac{\mathrm{\pi }}{2}\right)}-\tan \left(\theta \right)}}$

Here the numerator and denominator are divided by the $\tan \frac{\mathrm{\pi }}{2}$which givesright-side expression

So this step is justified

Step 4. Justification of step

Consider the step

$\frac{{\displaystyle \frac{\tan \left(\theta \right)}{\tan \left(\frac{\mathrm{\pi }}{2}\right)}+1}}{{\displaystyle \frac{1}{\tan \left(\frac{\mathrm{\pi }}{2}\right)}-\tan \left(\theta \right)}}=\frac{0+1}{0-\tan \left(\theta \right)}$

In this step, on the left-hand side, the value of$\tan \frac{\mathrm{\pi }}{2}$is undefined which make the left-hand side expression undefined

so this step is not justified and further derivation is not possible.

Step 5. Correct derivation

The correct derivation is

$$\begin{gathered} \tan \left(\theta +\frac{\mathrm{\pi }}{2}\right)=\frac{\sin \left(\theta +\frac{\mathrm{\pi }}{2}\right)}{\cos \left(\theta +\frac{\mathrm{\pi }}{2}\right)} \\ =\frac{\sin \theta \cos \frac{\mathrm{\pi }}{2}+\cos \theta \sin \frac{\mathrm{\pi }}{2}}{\cos \theta \cos \frac{\mathrm{\pi }}{2}-\sin \theta \sin \frac{\mathrm{\pi }}{2}} \\ =\frac{\sin \theta \left(0\right)+\cos \theta \left(1\right)}{\cos \theta \left(0\right)-\sin \theta \left(1\right)} \\ =\frac{0+\cos \theta }{0-\sin \theta } \\ =cot\theta \end{gathered}$$