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I am tasked with proving the this equality $$\frac{\cos(x)}{1 - \tan(x)} + \frac{\sin(x)}{1 - \cot(x)} = \sin(x) + \cos(x).$$ I've spent hours on it and no matter how much algebra I do, I can't figure it out. Can some one please help me?

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    $\begingroup$ Write $\tan(x) = \sin(x)/ \cos(x)$ etc and use $a^2-b^2 = (a+b)(a-b)$ $\endgroup$ – samjoe Jan 7 '18 at 13:46
  • $\begingroup$ Please improve your format, and show your work $\endgroup$ – Caffeine Jan 7 '18 at 13:53
  • $\begingroup$ Many answers seem to skip over this, $LHS$ is not well defined at many points while $RHS$ is well defined for all real $x$. $\endgroup$ – samjoe Jan 7 '18 at 14:40
  • $\begingroup$ HINT $$ \frac {c^2}{c-s} + \frac {s^2}{s-c} = \frac {c^2-s^2}{c-s} $$ $\endgroup$ – Narasimham Jan 7 '18 at 17:39
  • $\begingroup$ @Narasimham I think so too. $\endgroup$ – zwim Jan 8 '18 at 11:38
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Note that $\tan(x)=\sin(x)/\cos(x)=1/\cot(x)$ and $$\frac{\cos(x)}{1 - \tan(x)}+ \frac{\sin(x)}{1 -\cot(x)}= \frac{\cos^2(x)}{\cos(x) - \sin(x)}+ \frac{\sin^2(x)}{\sin(x) -\cos(x)}=\frac{\sin^2(x)-\cos^2(x)}{\sin(x) -\cos(x)}.$$ Can you take it from here?

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You said you spent hours on this problem, it is possible you get confused about all these trigonometric functions. One suggestion could be to replace $\cos(x)$ by $c$ and $\sin(x)$ by $s$.

If this $\quad\dfrac{c}{1-\frac sc}+\dfrac{s}{1-\frac cs}=\dfrac{c^2}{c-s}+\dfrac{s^2}{s-c}=\dfrac{c^2-s^2}{c-s}=\dfrac{(c-s)(c+s)}{c-s}=c+s$

is making more sense to you, even though it's just Robert.Z's answer written with simpler symbols, then go for this sort of reduction when you work with trigonometric formulas.

Myself for instance I always write $c^2+s^2=1$ instead of $\cos^2(x)+\sin^2(x)=1$, I find it easier to remember.

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  • $\begingroup$ Nice you like my trig shortcut cut notation. Also I use $t$ for tangent. Maybe no harm done when using with single independent variable. $\endgroup$ – Narasimham Jan 9 '18 at 9:06
  • $\begingroup$ @Narasimham this was ironic... You just wrote the same I did but 2h later. $\endgroup$ – zwim Jan 9 '18 at 9:23
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$1- cot(x)=\frac{ sin(x) -cos(x)}{sin(x)}$

$1- tan(x)=\frac{ cos(x) - sin(x)}{cos(x)}$

Then we have

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

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

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note that $$\frac{\cos(x)}{1-\tan(x)}+\frac{\sin(x)}{1-\cot(x)}-\sin(x)-\cos(x)=-\cos \left( x \right) \tan \left( x \right) \cot \left( x \right) - \tan \left( x \right) \sin \left( x \right) \cot \left( x \right) + \cos \left( x \right) \tan \left( x \right) +\sin \left( x \right) \cot \left( x \right) $$ and note that $$\tan(x)\cot(x)=1,\cos(x)\tan(x)=\sin(x),\sin(x)\cot(x)=\cos(x)$$

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