# Proving the identity $\frac{2\tan x}{1-\tan^2x}+\frac1{2\cos^2x-1} = \frac{\cos x+\sin x}{\cos x-\sin x}$

Prove $$\frac{2\tan x}{1-\tan^2x}+\frac1{2\cos^2x-1} = \frac{\cos x+\sin x}{\cos x-\sin x}$$

I know how to solve it, yet I can't!

first I join fractions (Easy)

then I "express" tans in sines and cosines

after it everything turns black!

With due recognition to the excellent solutions provided earlier, the solution below is an experiment in typography and trigonometric abbreviation.

Putting $s=\sin x, c=\cos x, t=\tan x$ and noting that $s/c=t$ and $c^2+s^2=1$, we have

$$\frac{c+s}{c-s}=\frac{(c+s)\color{blue}{^2}}{(c-s)\color{blue}{(c+s)}}=\frac{\overbrace{c^2+s^2}^1+2cs}{\underbrace{c^2-s^2}_{2c^2-1}}=\frac 1{2c^2-1}+\frac{2cs\qquad\color{green}{\div c^2}}{(c^2-s^2)\color{green}{\div c^2}}=\frac {2t}{1-t^2}+\frac 1{2c^2-1}$$

LHS $=\tan2x+\sec2x=\dfrac{\sin2x+1}{\cos2x}=\dfrac{(\cos x+\sin x)^2}{(\cos x+\sin x)(\cos x-\sin x)}=?$

OR

LHS=$\dfrac{2\sin x\cos x}{\cos^2x-\sin^2x}+\dfrac1{2\cos^2x-(\cos^2x+\sin^2x)}=\dfrac{\cos^2x+\sin^2x+2\sin x\cos x}{(\cos x+\sin x)(\cos x-\sin x)}=?$

• You are the fastest in the world !! Cheers (long time no speak) – Claude Leibovici Apr 18 '15 at 14:55
• @ClaudeLeibovici, Thanks. I'm an Introvert:) – lab bhattacharjee Apr 18 '15 at 14:58
• @ClaudeLeibovici I agree...I'm always astonished with how swift answers are produced. Reflective of two things really: 1) your intelligence in being able to solve the problem quickly and 2) your ability to typeset well. Well done! – Daniel W. Farlow Apr 18 '15 at 15:31
• @MagicMan. So, obviously, I don't satisfy any of these criteria ! Cheers :-) – Claude Leibovici Apr 18 '15 at 16:27
• @lab bhattacharjee its in which way is 2tanx/1-tan^2x + 1/2cos2(x)−1 equal to tan2x+sec2x? – Black Crescent Apr 18 '15 at 16:27

$$\frac { 2\tan(x)} { 1- \tan^2 (x) } + \frac 1 {2\cos^2(x)-1}$$

observe that $$\frac { 2\tan(x)} { 1- \tan^2 (x) } = (2\tan(x))\div(1- \frac{\sin^2(x)}{\cos^2(x)}) = (2\tan(x))\div \frac{\cos^2(x) - \sin^2(x)}{\cos^2(x)} = \frac{2\tan(x)\cos^2(x)}{\cos^2(x)-\sin^2(x)}$$ Note that $$2\cos^2(x)-1 = 2\cos^2(x) - \sin^2(x) - \cos^2(x) = \cos^2(x)-\sin^2(x)$$ So, $$\frac { 2\tan(x)} { 1- \tan^2 (x) } + \frac 1 {2\cos^2(x)-1} = \frac{2\tan(x)\cos^2(x)+1}{\cos^2(x)-\sin^2(x)} = \frac{2\sin(x)\cos(x)+1}{\cos^2(x)-\sin^2(x)}$$ Then note that: $2\sin(x)\cos(x)+1=2\sin(x)\cos(x)+1+\sin^2(x)+\cos^2(x) = (\cos(x)+\sin(x))^2$

Hence $$\frac { 2\tan(x)} { 1- \tan^2 (x) } + \frac 1 {2\cos^2(x)-1} = \frac{2\sin(x)\cos(x)+1}{\cos^2(x)-\sin^2(x)} = \frac {(\cos(x)+\sin(x))^2}{(\cos(x)-\sin(x))(\cos(x)+\sin(x))}= \frac {\cos(x)+\sin(x)}{\cos(x)-\sin(x)}$$

which is RHS

• You might want to try using a backslash before a trigo function for proper typesetting. – hypergeometric Apr 18 '15 at 16:18
• I still dont understand..., @asosnovsky how did the innocent looking 2tan(x)/1−tan2(x)+1/2cos2(x)−1 turn to 2tan(x)*cos^2(x)/cos^2(x)−sin^2(x)+1/2cos^2(x)−sin^2(x)−cos^2(x)? – Black Crescent Apr 18 '15 at 16:25
• because $1-\tan^2(x) = 1- \frac{ \sin^2(x) } { \cos^2(x) } = \frac { \cos^2(x) - \sin^2(x) } { \cos^2(x) }$ – asosnovsky Apr 18 '15 at 16:27
• @asosnovsky if it isnt too much trouble can you please detail the solution a bit further... – Black Crescent Apr 18 '15 at 16:31
• @Unknown I did, more detail than that is excessive – asosnovsky Apr 18 '15 at 16:43

I know there are already many solutions, but I am giving a very easy one. Please see!
$\frac { 2\tan(x)} { 1- \tan^2 (x) } + \frac 1 {2\cos^2(x)-1}$
The thing that should jump out at you is that the first term on the RHS is equal to $\tan(2x)$ and the second one is equal to $\frac{1}{\cos(2x)}$.
Now-
$\frac{\sin(2x)}{cos(2x)}+\frac{1}{\cos(2x)}$
$=\frac{2\sin(x)\cos(x)+1}{cos(2x)}$
$=\frac{(\sin(x)+\cos(x))^2}{\cos^2(x)-\sin^2(x)}$ (Because $1=\cos^2x+\sin^2x$)
$=\frac{\cos(x)+\sin(x)}{\cos(x)-\sin(x)}=RHS$

My basic strategy is $$LHS = LHS\cdot\color{blue}{\frac{1}{RHS}}\cdot \color{green}{RHS}=RHS$$ which is true if I can show that $$\frac{LHS}{RHS}=1$$

In Particular \begin{align} \frac{2\tan x}{1-\tan^2 x}+\frac{1}{2\cos^2 x -1}&=\bigg(\frac{2\tan x}{1-\tan^2 x}+\frac{1}{2\cos^2 x -1}\bigg)\cdot\color{blue}{\frac{\cos x - \sin x}{\cos x + \sin x}}\cdot \color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ &=\bigg(\frac{\cos^2 x}{\cos^2 x}\cdot\frac{2\tan x}{1-\tan^2 x}+\frac{1}{2\cos^2 x -1}\bigg)\cdot\color{blue}{\frac{\cos x - \sin x}{\cos x + \sin x}}\cdot \color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ &=\bigg(\frac{2\sin x\cos x}{\cos^2 x -\sin^2 x}+\frac{1}{\cos^2 x -1+\cos^2 x}\bigg)\cdot\color{blue}{\frac{\cos x - \sin x}{\cos x + \sin x}}\cdot \color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ &=\bigg(\frac{2\sin x\cos x + 1}{\cos^2 x -\sin^2 x}\bigg)\cdot\color{blue}{\frac{\cos x - \sin x}{\cos x + \sin x}}\cdot \color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ &=\bigg(\frac{2\sin x\cos x + 1}{(\cos x +\sin x)(\cos x -\sin x)}\bigg)\cdot\color{blue}{\frac{\cos x - \sin x}{\cos x + \sin x}}\cdot \color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ &=\bigg(\frac{2\sin x\cos x + 1}{\cos x +\sin x}\bigg)\cdot\color{blue}{\frac{1}{\cos x + \sin x}}\cdot \color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ &=\bigg(\frac{2\sin x\cos x + 1}{(\cos x +\sin x)^2}\bigg)\cdot \color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ &=\bigg(\frac{2\sin x\cos x + 1}{\cos^2 x + 2\sin x\cos x +\sin^2 x}\bigg)\cdot \color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ &=\bigg(\frac{2\sin x\cos x + 1}{2\sin x\cos x +1}\bigg)\cdot \color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ &=\color{green}{\frac{\cos x + \sin x}{\cos x - \sin x}}\\ \end{align}