Question

Establish the identity

$\frac{\tan \theta -cot\theta }{\tan \theta +cot\theta }=1-2{\cos }^2\theta$

Short answer

By converting tan and cot functions into sin and cos we get$\frac{\tan \theta -cot\theta }{\tan \theta +cot\theta }=1-2{\cos }^2\theta$

Step-by-step solution

Step 1. Given information     

$\frac{\tan \theta -cot\theta }{\tan \theta +cot\theta }=1-2{\cos }^2\theta$

Step 2. Rewriting the given expression 

Take LHS,$\frac{\tan \theta -cot\theta }{\tan \theta +cot\theta }$

$$\begin{gathered} \frac{{\displaystyle \frac{\sin \theta }{\cos \theta }}-{\displaystyle \frac{\cos \theta }{\sin \theta }}}{{\displaystyle \frac{\sin \theta }{\cos \theta }}+{\displaystyle \frac{\cos \theta }{\sin \theta }}}=\frac{{\sin }^2\theta -{\cos }^2\theta }{{\sin }^2\theta +{\cos }^2\theta }As{\sin }^2\theta +{\cos }^2\theta =1 \\ \frac{\tan \theta -cot\theta }{\tan \theta +cot\theta }={\sin }^2\theta -{\cos }^2\theta \end{gathered}$$

Step 3. Calculations 

Now,

$$\begin{gathered} \frac{\tan \theta -cot\theta }{\tan \theta +cot\theta }={\sin }^2\theta -{\cos }^2\theta \\ \frac{\tan \theta -cot\theta }{\tan \theta +cot\theta }=1-{\cos }^2\theta -{\cos }^{2}1=1-2{\cos }^2\theta \\ As{\sin }^2\theta +{\cos }^2\theta =1 \\ {\sin }^2\theta =1-{\cos }^2\theta \\ \frac{\tan \theta -cot\theta }{\tan \theta +cot\theta }=1-2{\cos }^2\theta \end{gathered}$$

LHS=RHS

Hence identity is established.