With this transformation problem, I need to transform cotxcosx + sinx into cscx and I've been at this for a while and would like to know how to do it correctly.
Hello, TLH!
Did you make any any substitutions?
Like $\displaystyle \frac{\cos x}{\sin x}$ for $\displaystyle \cot x$ ?
$\displaystyle \text{Show that: }\:\cot x\cos x + \sin x \:=\: \csc x$
We have: .$\displaystyle \frac{\cos x}{\sin x}\cdot\cos x} + \sin x \;=\;\frac{\cos^2\!x}{\sin x} + \sin x$
Get a common denominator: .$\displaystyle \frac{\cos^2\!x}{\sin x} + \frac{\sin x}{1}\cdot\frac{\sin x}{\sin x} \;=\;\frac{\cos^2\!x}{\sin x} + \frac{\sin^2\!x}{\sin x}$
Add: .$\displaystyle \frac{\overbrace{\cos^2\!x + \sin^2\!x}^{\text{This is 1}}}{\sin x} \;=\;\frac{1}{\sin x} \;=\;\csc x$
... Wow, the way to solving the proof was staring right in front of me. Though, that's the case whenever I'm stumped and ask help for these types of questions.
Thanks.
And yes, I did make substitutions. I did many things, except for that.