Question

Hi, I was absent today in class and missed the whole lesson, this is not a grade exam this is a homework that my teacher assigned us to do, can you please explain how to do it much appreciated it I tried this question by my own but it was wrong

Answer 1

First, let's factor the term "tan(x)" on the left side:

$\begin{gathered} \tan x\csc ^2x-\tan x=\cot x \\ \tan x(\csc ^2x-1)=\cot x \end{gathered}$

Then, we use the following trigonometrical identity:

$\begin{gathered} \csc ^2x-1=\cot ^2x \\ \\ \text{proof:} \\ \csc ^2x-1=\frac{1}{\sin^2x}-1=\frac{1-\sin^2x}{\sin^2x}=\frac{\cos^2x}{\sin^2x}=\cot ^2x \end{gathered}$

So we have:

$\begin{gathered} \tan x(\csc ^2x-1)=\cot x \\ \tan x\cdot\cot ^2x=\cot x \end{gathered}$

Now we use the fact that the cotangent is 1 over the tangent:

$\begin{gathered} \tan x\cdot\cot ^2x=\cot x \\ \tan x\cdot\frac{1}{\tan^2x}=\cot x \\ \frac{1}{\tan x}=\cot x \\ \cot x=\cot x \end{gathered}$