# Show that $\frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}= 2 \csc x$

Verify the following identity: $$\frac{\tan x}{1+\sec x}+\frac{1+\sec x}{\tan x}= 2 \csc x$$

-
Consider what you must do in order to add the two fractions $\frac{\tan x}{1+\sec x} + \frac{1+\sec x}{\tan x}$ , that is, after putting them over a common denominator. You will also want to use the variant form of the Pythagorean Identity, $1 + \tan^2 x = \sec^2 x$. Beyond that, you will need to do some algebra and find a common factor in the numerator and denominator to cancel. – RecklessReckoner Apr 15 '13 at 1:51
I got this far.. (sin^2x/cos^2x)+1+(2/cosx)+(1/cos^2x)+(1/sinx) – user72588 Apr 15 '13 at 2:07
I was thinking more along the lines of $\frac{(\tan x \cdot \tan x) + (1 + \sec x)(1 + \sec x)}{(1 + \sec x)(\tan x)}$ by adding fractions. Multiply out the numerator and simplify, but leave the denominator alone for now. – RecklessReckoner Apr 15 '13 at 2:18
It might be easier to multiply the numerator and denominator of the first fraction by $\sec x-1$. – Mike Apr 15 '13 at 3:10

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 $$\frac{1+\sec x}{\tan x}=\frac{1+\frac1{\cos x}}{\frac{\sin x}{\cos x}}=\frac{1+\cos x}{\sin x}$$

-
In Method 1, representing $\tan x$ as $\frac{\sec x}{\csc x}$ would be much cleaner – John Joy Aug 1 '14 at 17:17
@JohnJoy, Please find the edited version – lab bhattacharjee Aug 1 '14 at 18:07

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.

-
I got this far.. (sin^2x/cos^2x)+1+(2/cosx)+(1/cos^2x)+(1/sinx) – user72588 Apr 15 '13 at 1:54
@User: I suppose you should find a common denominator. – mixedmath Apr 15 '13 at 2:17
You're going to need to write out your steps -- I think something is off in this result. – RecklessReckoner Apr 15 '13 at 2:22

$$\dfrac{\tan x}{1+\sec x}+\dfrac{1+\sec x}{\tan x}$$ $$\dfrac{\tan^2 x+(1+\sec x)^2}{(1+\sec x)\tan x}$$ $$\dfrac{\tan^2 x+1+\sec^2x+2\sec x}{(1+\sec x)\tan x}$$ $$\dfrac{\sec^2x+\sec^2x+2\sec x}{(1+\sec x)\tan x}$$ $$\dfrac{2\sec x(\sec x+1)}{(1+\sec x)\tan x}$$ $$\dfrac{2\sec x}{\tan x}$$ $$\dfrac{\frac {2}{\cos x}}{\frac{\sin x}{\cos x}}$$ $${\dfrac {2}{\cos x}}\times\dfrac{\cos x}{\sin x}$$ $$\dfrac 2 {\sin x}$$ $$2\csc x$$

-