# Is there an elegant way to simplify $\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}$

I wonder how to solve this equation: $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ in an elegant/shorter way. My way: $$\frac{\sin(x+20^{\circ })-\sin(x+20^{\circ })\cdot cos(x+20^{\circ})}{\sin(x+20^{\circ })+\sin(x+20^{\circ })\cdot \cos(x+20^{\circ})}=2[1-(\cos(x+20^{\circ}))]$$ so $$1-\cos(x+20^{\circ})=2[1-\cos(x+20^{\circ})][1+\cos(x+20^{\circ})]$$ which gives $$\cos(x+20^{\circ})=1\Rightarrow x=-20^{\circ}+360^{\circ}k\\\cos(x+20^{\circ})=-0.5\Rightarrow x=140^{\circ}+360^{\circ}k,\, \, x=-100^{\circ}+360^{\circ}k$$ Thanks.

Setting $\displaystyle\frac x2+10^\circ=y\iff x+20^\circ=2y$

$$\frac{\tan2y-\sin2y}{\tan2y+\sin2y}=\frac{\sin2y\cdot\dfrac{1-\cos2y}{\cos2y}}{\sin2y\cdot\dfrac{1+\cos2y}{\cos2y}}=\frac{\cos2y\sin2y(1-\cos2y)}{\cos2y\sin2y(1+\cos2y)}=\tan^2y$$ if $\cos2y\sin2y\ne0$

Now $\cos2y\sin2y=0\iff\sin4y=2\sin2y\cos2y=0\implies4y=n\cdot180^\circ$ where $n$ is any integer

So, we need $\displaystyle y\ne n\cdot45^\circ\ \ \ \ (1)$ and we have $\displaystyle\tan^2y=4\sin^2y,$

$\displaystyle\implies\sin^2y=4\sin^2y\cos^2y\iff\sin^2y(1-4\cos^2y)=0$

If $\displaystyle\sin^2y=0,\sin y=0\implies y=m\cdot180^\circ$ which is inadmissible by $(1)$

So, $\displaystyle1-4\cos^2y=0\iff\cos^2y=\frac14\iff\cos2y=2\cos^2y-1=-\frac12=\cos120^\circ$

$\displaystyle\implies2y=360^\circ\cdot r\pm120^\circ\iff y=180^\circ\cdot r\pm60^\circ$ where $r$ is any integer

So, $x=2y-20=\cdots$

• can you please check my answer? – shaurya gupta May 12 '14 at 7:22
• @shauryagupta, I think you proceed a little further to discard the inadmissible value – lab bhattacharjee May 12 '14 at 12:48
• @labbhattacharjee - so $x=2y-20^\circ=-20^\circ+360^\circ\cdot r\pm120^\circ$? – aric May 12 '14 at 21:42
• @aric, Any doubt? Please validate the values – lab bhattacharjee May 13 '14 at 2:47

$$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=\\\frac{\frac{\sin(x+20^{\circ })}{\cos(x+20^{\circ })}-\sin(x+20^{\circ })}{\frac{\sin(x+20^{\circ })}{\cos(x+20^{\circ })}+\sin(x+20^{\circ })}=\\\frac{\sin(x+20^{\circ })}{\sin(x+20^{\circ })}\frac{\frac{1}{\cos(x+20^{\circ })}-1}{\frac{1}{\cos(x+20^{\circ })}+1}=\\\frac{\sin(x+20^{\circ })}{\sin(x+20^{\circ })}\frac{\cos(x+20^{\circ })}{\cos(x+20^{\circ })}\frac{1-\cos(x+20^{\circ })}{1+\cos(x+20^{\circ })}=\\\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\frac{2\sin^2(\frac{x}{2}+10^{\circ })}{2\cos^2(\frac{x}{2}+10^{\circ })}=\\\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\tan^2\left(\frac{x}{2}+10^{\circ }\right)$$ Thus you can rewrite $$\frac{\tan(x+20^{\circ })-\sin(x+20^{\circ })}{\tan(x+20^{\circ })+\sin(x+20^{\circ })}=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)$$ as $$\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\tan^2\left(\frac{x}{2}+10^{\circ }\right)=4\sin^{2}\left(\frac{x}{2}+10^{\circ }\right)\\\frac{\tan(x+20^{\circ })}{\tan(x+20^{\circ })}\frac{1}{\cos^2(\frac{x}{2}+10^{\circ })}=\frac{4}{\cos^2(\frac{x}{2}+10^{\circ })}\\\frac{\tan(x+20^{\circ })}{\cos^2(\frac{x}{2}+10^{\circ })}=\frac{4\tan(x+20^{\circ })}{\cos^2(\frac{x}{2}+10^{\circ })}\\\frac{3\tan(x+20^{\circ })}{\cos^2(\frac{x}{2}+10^{\circ })}=0\Rightarrow\\\tan(x+20^{\circ })=0\\x+20^{\circ }=0^{\circ }\\x=-20^{\circ }+k\pi^{\circ }\qquad k\in\Bbb{Z}$$

• why is $4sin^{2}(\frac{x}{2}+10^{\circ})=\frac{4}{cos^{2}(\frac{x}{2}+10^{\circ})}$ – aric May 11 '14 at 18:55
• @Foga- also you got: $1=4$ so i think something is wrong in your solution. – aric May 11 '14 at 19:08
• Nothing! You obtain $1=4$ by dividing by $0$... – sirfoga May 11 '14 at 19:39
• @Foga, But $x=180^\circ n-20^\circ$ will make the LHS $\dfrac00$ – lab bhattacharjee May 12 '14 at 6:55
• I've edited the answer to make as clear as possible... – sirfoga May 12 '14 at 11:09

$4sin^2(\frac{x + 20^\circ}2) = 4\cdot1/2\cdot(1-\cos(x+20^\circ)) = 2-2\cos(x+20^\circ)$

Applying componendo and dividendo on $\frac{tg(x+20^{\circ })-sin(x+20^{\circ })}{tg(x+20^{\circ })+sin(x+20^{\circ })}=4sin^{2}(\frac{x}{2}+10^{\circ })$ ,

$-\sec(x+20^\circ) = \frac{2-2\cos(x+20^\circ)+1}{2-2\cos(x+20^\circ) -1}$, $cos(x+20^\circ)\ne1/2$
$-\sec(x+20^\circ)\cdot(1-{2\cos(x+20^\circ)}) = {3-2\cos(x+20^\circ)}$
$-\sec(x+20^\circ)+2=3-2\cos(x+20^\circ)$
$2cos(x+20^\circ)-\sec(x+20^\circ)-1=0$
$2cos^2(x+20^\circ)-cos(x+20^\circ)-1=0$
Can you solve it further?