Chapter 7, Problem 16CT

To determine

To prove: The identity $\frac{\tan \theta -\cot \theta }{\tan \theta +\cot \theta }=1-2{\cos }^2\theta$.

Explanation of Solution

Given information:

Instruction is to establish the identity $\frac{\tan \theta -\cot \theta }{\tan \theta +\cot \theta }=1-2{\cos }^2\theta$

Formula used:

1) $\cot \theta =\frac{\cos \theta }{\sin \theta }$

2) $\tan \left(\theta \right)=\frac{\sin \theta }{\cos \theta }$

3) Pythagorean identity: ${\sin }^2\theta +{\cos }^2\theta =1$

Proof:

Consider L.H.S. of the identity,

$\frac{\tan \theta -\cot \theta }{\tan \theta +\cot \theta }$

By using $\cot \theta =\frac{\cos \theta }{\sin \theta }$ and $\tan \left(\theta \right)=\frac{\sin \theta }{\cos \theta }$,

$\Rightarrow \frac{\frac{\sin \theta }{\cos \theta }-\frac{\cos \theta }{\sin \theta }}{\frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta }}$

By cross multiplication,

$\Rightarrow \frac{\frac{{\sin }^2\theta -{\cos }^2\theta }{\sin \theta \cos \theta }}{\frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta }}$

Simplify by cancelling the common term from the numerator and denominator,

$\Rightarrow \frac{{\sin }^2\theta -{\cos }^2\theta }{{\sin }^2\theta +{\cos }^2\theta }$

By using Pythagorean identity ${\sin }^2\theta +{\cos }^2\theta =1$,

$\Rightarrow \frac{{\sin }^2\theta -{\cos }^2\theta }{1}$

$={\sin }^2\theta -{\cos }^2\theta$

Again, by using Pythagorean identity ${\sin }^2\theta =1-{\cos }^2\theta$,

$\Rightarrow 1-{\cos }^2\theta -{\cos }^2\theta$

$=1-2{\cos }^2\theta$

= R.H.S.

Hence, the identity $\frac{\tan \theta -\cot \theta }{\tan \theta +\cot \theta }=1-2{\cos }^2\theta$ is proved