Verifying identities, trigonometry I am completely stuck.  I cannot come up with a proof for this identity.  I am ready to pull my hair out.  $$\sin(x) \left(\tan(x)+ \frac 1{\tan(x)}\right) = \sec(x)$$
 A: Hint: Remember $\tan(x) = \frac{\sin(x)}{\cos(x)}$ and $1 + \tan^2(x) = \sec^2(x)$.
Edit to reflect full answer:
Because we know that both $\tan(x)$ and $\sec (x)$ can be written in terms of $\sin(x)$ and $\cos(x)$, we will rewrite them. Remembering:
$$\tan(x) = \frac{\sin(x)}{\cos(x)} \ \ \mathrm{and } \ \ 1 + \tan^2(x) = \sec^2(x)\ \ .$$
We have our original formula rewritten as :
\begin{align}
\sin(x) \left(\tan(x)+ \frac 1{\tan(x)}\right) &=\sin(x) \left(\tan(x)+ \tan^{-1}(x)\right) \\
&=\sin(x) \left(\frac{\sin(x)}{\cos(x)}+ \left(\frac{\sin(x)}{\cos(x)}\right)^{-1}\right)\\
&= \sin(x) \left(\frac{\sin(x)}{\cos(x)}+ \frac{\cos(x)}{\sin(x)}\right)\\
&=\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right)\ .
\end{align}
From here there are two techniques we can use to prove this identity. 
Technique 1: 
In this technique we try to put our identity into the form $1 + \tan^2(x) = \sec^2(x)$, then get to $\sec(x)$
We can see that the only thing necessary for the $\frac{\sin^2(x)}{\cos(x)}$ term to be equal to $\tan^2(x)$ it to divide by $\cos(x)$, and then for the $\cos(x)$ term to equal $1$ we also need to divide by $\cos(x)$. This means we need to multiply our whole expression by $\frac{1}{\cos(x)}$, but because we can not change our expression, we must multiply by one $1$ (because one is the identity for multiplication), therefore we must multiply by $\frac{\cos(x)}{\cos(x)}$.
Now we preform the multiplication:
$$\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right) \cdot \frac{\cos(x)}{\cos(x)} = \left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right) \cdot \frac{1}{\cos(x)} \cdot \cos(x) $$
We can distribute the $ \frac{1}{\cos(x)}$ to get:
\begin{align}
\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right) \cdot \frac{1}{\cos(x)} \cdot \cos(x) &= \left(\frac{\sin^2(x)}{\cos^2(x)}+ \frac{\cos(x)}{\cos(x)}\right) \cdot \cos(x)\\
&= \left(\tan^2(x)+ 1\right) \cdot \cos(x) \ \ .
\end{align}
Now we can use the identity $1 + \tan^2(x) = \sec^2(x)$:
$$\left(\tan^2(x)+ 1\right) \cdot \cos(x) = \sec^2(x) \cdot \cos(x)$$
and because $ \sec(x) = \frac{1}{\cos(x)}$:
\begin{align}
\sec^2(x) \cdot \cos(x) &= \left(\frac{1}{\cos(x)} \right)^2 \cdot \cos(x)\\
&= \frac{1}{\cos^2(x)} \cdot \cos(x)\\
&= \frac{1}{\cos(x)}\\
&= \sec(x) 
\end{align}
As desired. 
Technique 2: 
Here again we start with the expression $\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right)$. Now $\sec(x)$ is a single term, and our expression has two terms, so we will try to condense our expression down to one term by putting the fractions under common denominators: 
\begin{align}
\left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\right) &=  \left(\frac{\sin^2(x)}{\cos(x)}+ \cos(x)\cdot \frac{cos(x)}{\cos(x)} \right)\\
&= \left(\frac{\sin^2(x)}{\cos(x)}+ \frac{cos^2(x)}{\cos(x)} \right)\\
&= \left(\frac{\sin^2(x) + cos^2(x)}{\cos(x)} \right)\ .
\end{align}
We now can use the Pythagorean identity ($\sin^2(x) + \cos^2(x) = 1$):
\begin{align}
\left(\frac{\sin^2(x) + cos^2(x)}{\cos(x)} \right) &= \frac{1}{\cos(x)} \\
&= \sec(x)
\end{align}
as desired.
Remarks:
While both of the techniques are valid proofs, the second technique is (in my personal opinion) more desirable because it is both shorter and it is more intuitive. While both proofs use the Pythagorean identity (the first uses an indirect Pythagorean identity, attained by dividing the Pythagorean identity by $\cos^2(x)$), the second uses it without needing to do a large amount of manipulation to attain the applicable form. If I were presented with this problem on an exam I would most likely use the second technique (although the first came to mind first, as reflected in my hint), but like I said, either is valid. 
A: This is actually very simple. So, here's how it goes:
We have to prove $sin(x)[$tan(x)+($\frac{1}{tan(x)}$)]=$sec(x)$
Now replacing $tan(x)$ = $\frac{sin(x)}{cos(x)}$ and $\frac{1}{tan(x)}$ = $\frac{cos(x)}{sin(x)}$ in the $LHS$, we get: 
$LHS$ $=$ $sin(x)$[($\frac{sin(x)}{cos(x)}$) $+$ ($\frac{cos(x)}{sin(x)}$)] 
$=$ $ sin(x)$[$\frac{(sin^2 (x) + cos^2 (x))}{sin(x).cos(x)}$] 
Here, $sin^2 (x) + cos^2 (x)$ is 1 and $sin(x)$ gets cancelled, which gives us: 
=> $\frac{1}{cos(x)}$ = $sec(x)$ 
Hence Proved.
A: $$\frac{1}{\tan x}+\tan x = \frac{1+\tan^2x}{tanx}$$
now $1+\tan^2x=\sec^2 x$ and $tanx =\frac{\sin x}{\cos x}$
