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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!

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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

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

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

OR

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

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  • $\begingroup$ You are the fastest in the world !! Cheers (long time no speak) $\endgroup$ – Claude Leibovici Apr 18 '15 at 14:55
  • $\begingroup$ @ClaudeLeibovici, Thanks. I'm an Introvert:) $\endgroup$ – lab bhattacharjee Apr 18 '15 at 14:58
  • $\begingroup$ @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! $\endgroup$ – Daniel W. Farlow Apr 18 '15 at 15:31
  • $\begingroup$ @MagicMan. So, obviously, I don't satisfy any of these criteria ! Cheers :-) $\endgroup$ – Claude Leibovici Apr 18 '15 at 16:27
  • $\begingroup$ @lab bhattacharjee its in which way is 2tanx/1-tan^2x + 1/2cos2(x)−1 equal to tan2x+sec2x? $\endgroup$ – Black Crescent Apr 18 '15 at 16:27
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Easy, start with leftside

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

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

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

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

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