Hello jgv115 Originally Posted by

**jgv115** Thanks Grandpa, I really appreciate your help.

I have drawn this diagram up:

By using pythagoras we can get

$\displaystyle \cos^2 \theta + \sin^2 \theta = 1 $

But the book I have here says you can also get these:

$\displaystyle cot^2 \theta +1 = cosec^2 \theta $

and

$\displaystyle 1+\tan^2\theta = \sec^2 \theta $

i'm so confused by this. I have no idea how to get these. Could anyone provide assistance?

Thanks in advanced

Your book is quite correct (and, incidentally, it means that, theoretically, you hardly need cosine either, you can just about get by with sine, and $\displaystyle \cos^2\theta = 1 - \sin^2\theta$!)

The other two equations are very easy to derive, if you remember that $\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}$ and therefore that $\displaystyle \cot\theta = \frac{\cos\theta}{\sin\theta}$. So start with: $\displaystyle \cos^2\theta+\sin^2\theta = 1$

and divide both sides, first of all, by $\displaystyle \sin^2\theta$:$\displaystyle \frac{\cos^2\theta}{\sin^2\theta}+\frac{\sin^2\the ta}{\sin^2\theta}=\frac{1}{\sin^2\theta}$

$\displaystyle \Rightarrow \cot^2\theta+1 = \csc^2\theta$

Now go back and divide both sides by $\displaystyle \cos^2\theta$:$\displaystyle \frac{\cos^2\theta}{\cos^2\theta}+\frac{\sin^2\the ta}{\cos^2\theta}=\frac{1}{\cos^2\theta}$

$\displaystyle \Rightarrow 1 + \tan^2\theta = \sec^2\theta$

Grandad