Chapter 7.5, Problem 110AYU

To determine

Verify the following derivation and justify each step.

Answer to Problem 110AYU

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

Explanation of Solution

Given information:

Verify the following derivation and justify each step.

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

Calculation:

We have to prove.

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

Now consider the given function,

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

In this step sum formula for tangent identity $\tan \left(\alpha +\beta \right)=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$ is used where $\theta$ represents $\alpha$ and $\frac{\pi }{2}=\beta$

Hence this step is justified.

Now consider another part of equation to verification.

  $\frac{\tan \theta +\tan \frac{\pi }{2}}{1-\tan \theta \tan \frac{\pi }{2}}=\frac{\frac{\tan \theta }{\tan \frac{\pi }{2}}+1}{\frac{1}{\tan \frac{\pi }{2}}-\tan \theta }$

In above expression left hand side of numerator and denominator is divided by $\tan \frac{\pi }{2}$

Hence, this step is also verified.

Now consider,

  $\frac{\frac{\tan \theta }{\tan \frac{\pi }{2}}+1}{\frac{1}{\tan \frac{\pi }{2}}-\tan \theta }=\frac{0+1}{0-\tan \theta }$

This step is justified because $\tan \frac{\pi }{2}$ is undefined hence left hand side of expression is undefined.

Now consider,

  $\frac{0+1}{0-\tan \theta }=\frac{1}{-\tan \theta }$

This step is not justified, because identity should not be justified as $\frac{1}{-\tan \theta }$ turn to $-\cot \theta$

Now the correct proof of the identity $\tan \left(\theta +\frac{\pi }{2}\right)$

  $=-\cot \theta$ is,

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

Hence, $\boxed{\tan \left(\theta +\frac{\pi }{2}\right)=-\cot \theta }$ verified.