Prove that $\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$ = $\frac{1+\tan(a)}{\sec(a)}$ 

Prove that $\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$ = $\frac{1+\tan(a)}{\sec(a)}$


My attempt using the LHS 
$$\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$$
$$ \frac{2\sin(a)+\frac{1}{\cos(a)}}{1+\frac{\sin(a)}{\cos(a)}} $$ 
$$ \frac{\frac{2\sin(a)+1}{\cos{a}}}{\frac{\cos(a)+\sin(a)}{\cos(a)}} $$
$$ {\frac{2\sin(a)+1}{\cos{a}}} * {\frac{\cos(a)}{\cos(a)+sin(a)}} $$ 
$$ \frac{2\sin(a)+1}{\cos(a)+sin(a)}    $$
Now I am stuck...
 A: Hint: What does $(\cos a+ \sin a)^2$ simplify to?
A: As mentioned in the comments,looks like you made a calculation error.
Anyways heres the math-
$\frac{2\sin(a)+\sec(a)}{1+\tan(a)}$=$\frac{2\sin(a) \cos(a)+1}{{\cos(a)+\sin(a)}}$
Now knowing that ${1+\sin(2a)}={(\sin(a)+\cos(a))^2}$...the question is a piece of cake.
A: Although no method is foolproof, I've had few problems with this one.
Step 1: Take what you started with and multiply by 1.
$$\frac{2\sin a+\sec a}{1+\tan a}\cdot 1$$
Step 2: Split up 1 into what you want to end up with and its reciprocal.
$$\frac{2\sin a+\sec a}{1+\tan a}\cdot\bigg(\frac{\sec a}{1+\tan a}\cdot\frac{1+\tan a}{\sec a}\bigg)$$
Step 3: Group the reciprocal with what you started with.
$$\bigg(\frac{2\sin a+\sec a}{1+\tan a}\cdot\frac{\sec a}{1+\tan a}\bigg)\cdot\frac{1+\tan a}{\sec a}$$
Step 4: Show that what is in parentheses is equal to 1.
$$=\bigg(\frac{2\sin a\sec a+\sec^2 a}{(1+\tan a)^2}\bigg)\cdot\frac{1+\tan a}{\sec a}$$
$$=\bigg(\frac{2\tan a+(\tan^2 a+1)}{(1+\tan a)^2}\bigg)\cdot\frac{1+\tan a}{\sec a}$$
$$=\bigg(\frac{\tan^2 a+2\tan a+1}{(1+\tan a)^2}\bigg)\cdot\frac{1+\tan a}{\sec a}$$
$$=\bigg(\frac{(1+\tan a)^2}{(1+\tan a)^2}\bigg)\cdot\frac{1+\tan a}{\sec a}$$
$$=1\cdot\frac{1+\tan a}{\sec a}=\frac{1+\tan a}{\sec a}$$
A: Avoid cumbersome workarounds, just cross multiply
$$\frac{2\sin(a)+\sec(a)}{1+\tan(a)}=\frac{1+\tan(a)}{\sec(a)}$$
$$ 2 \tan a + \sec^2 a = 1 + 2 \tan a + \tan^{2} a $$ 
and that is correct!
A: Let $s=\sin(a)$, $c=\cos(a)$, $k=\sec(a) = 1/c$, $t=\tan(a)=s/c \;$ to eliminate visual clutter.
$$\frac{2\sin(a)+\sec(a)}{1+\tan(a)} = \frac{2s+k}{1+t} = \frac{2s+1/c}{1+s/c}=\frac{2sc+1}{c+s}=\frac{(c+s)^2}{c+s}= c+s \\= \frac{1+s/c}{1/c}=\frac{1+t}k = \frac{1+\tan(a)}{\sec(a)}$$
A: $$(1+\tan A)\cdot(1+\tan A)=1+\tan^2A+2\tan A=\sec^2A+2\tan A$$
$$\iff(1+\tan A)\cdot(1+\tan A)=\sec A(\sec A+2\sin A)$$
