katlynsbirds' question at Yahoo! Answers regarding inverse trigonometric identity

Staff member

MarkFL

Administrator
Staff member
Re: katlynsbirds' question at Yahoo! Answers regarding inverse trignometric identity

Hello katlynsbirds,

We are given to prove:

$$\displaystyle \cot^{-1}(x)=\sin^{-1}\left(\frac{1}{\sqrt{1+x^2}} \right)$$

Let's let $$\displaystyle \theta=\cot^{-1}(x)\,\therefore\,x=\cot(\theta)$$, and now please refer to this diagram:

We see that $$\displaystyle \cot(\theta)=\frac{x}{1}=x$$ and we can also see that:

$$\displaystyle \sin(\theta)=\frac{1}{\sqrt{1+x^2}}\,\therefore\, \theta=\sin^{-1}\left(\frac{1}{\sqrt{1+x^2}} \right)$$

and so we may conclude:

$$\displaystyle \theta=\cot^{-1}(x)=\sin^{-1}\left(\frac{1}{\sqrt{1+x^2}} \right)$$

Shown as desired.

To katlynsbirds and any other guests viewing this topic I invite and encourage you to post other trigonometry problems here in our Trigonometry forum.

Best Regards,

Mark.