Question

Explain why formula cannot be used to show that

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

Establish this identity by using formulas $\left(3a\right)$and$\left(3b\right)$.

Short answer

The equation$\tan \left(\frac{\mathrm{\pi }}{2}-\theta \right)=cot\theta$cannot be derived using the formula$\tan \left(\alpha +\beta \right)=\frac{\tan \left(\alpha \right)+\tan \left(\beta \right)}{1-\tan \left(\alpha \right)\tan \left(\beta \right)}$becouse this derivation will consist 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 formula is

$\tan \left(\alpha +\beta \right)=\frac{\tan \left(\alpha \right)+\tan \left(\beta \right)}{1-\tan \left(\alpha \right)\tan \left(\beta \right)}$

The equation we need to prove is

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

Step 2. Clarification

Derivation using formula $\tan \left(\alpha +\beta \right)=\frac{\tan \left(\alpha \right)+\tan \left(\beta \right)}{1-\tan \left(\alpha \right)\tan \left(\beta \right)}$

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

Not the expression consists of $\tan \frac{\mathrm{\pi }}{2}$

which is undefined so the further derivation is not possible

Step 3. correct derivation

Derivation of the equation

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