Proving Trig Identities: cos x (csc x - sec x) = cot x - 1

Sometimes you just don't see it.

Have you tried turning csc and sec into sin and cos?
 
Yeah I turned csc x into 1/sin x and sec x into 1/cos x and cot into cos x / sin x.
 
\(\displaystyle \L
\begin{array}{l}
\cos (x) \cdot \csc (x) = \frac{{\cos (x)}}{{\sin (x)}} = \cot (x) \\
\; \\
\cos (x) \cdot \sec (x) = \frac{{\cos (x)}}{{\cos (x)}} = 1 \\
\end{array}\)
 
Re: Proving Trig Identities: cos x (csc x - sec x) = cot x -

Hello, Verde77!

How is \(\displaystyle \cos x (\csc x\,-\,\sec x)\:=\:\cot x\,-\,1\,\) proved?
Click to expand...
First of all, do NOT work with both sides of the identity.
Try to make one side (usually the "messier" one) equal to the other side.

The left side is: \(\displaystyle \,\cos x(\csc x\,-\,\sec x) \;=\;\cos x\left(\frac{1}{\sin x}\,-\,\frac{1}{\cos x{\right)\)

Multiply: \(\displaystyle \;\cos x\,\cdot\,\frac{1}{\sin x}\:-\:\cos x\,\cdot\,\frac{1}{\cos x} \;\;= \;\;\frac{\cos x}{\sin x}\,-\,1\;\;=\;\;\cot x\,-\,1\;\) . . . There!
 
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