Show that $ \frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}= 2 \csc x$ Verify the following identity: 
$$ \frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}= 2 \csc x$$
 A: Method  $1:$
As
$\displaystyle\tan^2x=\sec^2x-1=(\sec x-1)(\sec x+1),\frac{\tan x}{\sec x-1}=\frac{\sec x+1}{\tan x}$
$$\implies\frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}=\frac{\sec x-1}{\tan x}+\frac{1+\sec x}{\tan x}=\frac{2\sec x}{\tan x}=\frac{\dfrac2{\cos x}}{\dfrac {\sin x}{\cos x}}=\frac2{\sin x}$$
Method  $2:$
$$\frac{\tan x}{1+\sec x}=\frac{\frac {\sin x}{\cos x}}{1+\frac1{\cos x}}=\frac{\sin x}{1+\cos x}=\frac{\sin x(1-\cos x)}{(1+\cos x)(1-\cos x)}=\frac{1-\cos x}{\sin x}\text{ as } \sin^2x=1-\cos^2x$$
and 
$$\frac{1+\sec x}{\tan x}=\frac{1+\frac1{\cos x}}{\frac{\sin x}{\cos x}}=\frac{1+\cos x}{\sin x}$$
A: If you get stuck, you convert everything to sine and cosine values. Conceivably, use the Pythagorean theorem.
If this doesn't work, post your attempt and we can guide you from there.
A: $$ \dfrac{\tan x}{1+\sec x}+\dfrac{1+\sec x}{\tan x}$$
$$\dfrac{\tan^2 x+(1+\sec x)^2}{(1+\sec x)\tan x}$$
$$\dfrac{\tan^2 x+1+\sec^2x+2\sec x}{(1+\sec x)\tan x}$$
$$\dfrac{\sec^2x+\sec^2x+2\sec x}{(1+\sec x)\tan x}$$
$$\dfrac{2\sec x(\sec x+1)}{(1+\sec x)\tan x}$$
$$\dfrac{2\sec x}{\tan x}$$
$$\dfrac{\frac {2}{\cos x}}{\frac{\sin x}{\cos x}}$$
$${\dfrac {2}{\cos x}}\times\dfrac{\cos x}{\sin x}$$
$$\dfrac 2 {\sin x}$$
$$2\csc x$$
