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=(1)/(\sin^2x)-1=(1-\sin^2x)/(\sin^2x)=(\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(1)/(\tan^2x)=\cot x \\ (1)/(\tan x)=\cot x \\ \cot x=\cot x \end{gathered}`