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This is a tricky trig problem I'm stuck with. The problem is asking me to simplify $$1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1}$$ to $\sin x\cos x$.


What I've been doing so far is trying to remove those $tan$ functions.
$$1-\left(\frac{\sin^2x\tan x-\cos^2}{\tan x+1}\right)$$ $$1-\left(\frac{\frac{\sin^3x-\cos^3x}{\cos x}}{\tan x+1}\right)$$ $$1-\left(\frac{\frac{\sin^3x-\cos^3x}{\cos x}}{\frac{\sin x}{\cos x}+1}\right)$$ I made this complicated. Is there a simple way to do this problem?

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  • $\begingroup$ MathJax hint: if you put a backslash before common functions you get the right font and spacing, so \sin x gives $\sin x$ instead of sin x which gives $sin x$ $\endgroup$ – Ross Millikan May 8 '18 at 14:04
  • $\begingroup$ Noted! Still learning how to use the MathJax syntax. $\endgroup$ – AugieJavax98 May 8 '18 at 14:06
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We have

$$1-\frac{\sin^2x\tan x}{\tan x+1}-\frac{\cos^2x}{\tan x+1} =1-\frac{\sin^2x\tan x+\cos^2 x}{\tan x+1}=1-\frac{\sin^3x+\cos^3 x}{\sin x + \cos x}=\\=\frac{\sin x(1-\sin^2x)+\cos x(1-\cos^2 x)}{\sin x + \cos x}=\frac{\sin x\cos^2x+\cos x\sin^2 x}{\sin x + \cos x}=\sin x \cos x$$

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  • $\begingroup$ I am confused with the last part of your working. Did you use another identity? $\endgroup$ – AugieJavax98 May 8 '18 at 14:50
  • $\begingroup$ Sorry I skip a step, that is $$\frac{\sin x\cos^2x+\cos x\sin^2 x}{\sin x + \cos x}=\frac{\sin x \cos x(\cos x+\sin x)}{\sin x + \cos x}=\sin x \cos x$$ $\endgroup$ – gimusi May 8 '18 at 14:51
  • $\begingroup$ Oh! I see! What tips can you give for tackling such trig problems? $\endgroup$ – AugieJavax98 May 8 '18 at 14:55
  • $\begingroup$ Try to learn the way to obtain and proof all the main trigonometric identities and do a lot of exercises on that! Do not try to memorize all the expressions but try to understand them and check them by simple cases. That's always a good reference en.wikipedia.org/wiki/List_of_trigonometric_identities Bye! $\endgroup$ – gimusi May 8 '18 at 15:00
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    $\begingroup$ Noted! I guess practice makes perfect. Thanks again $\endgroup$ – AugieJavax98 May 8 '18 at 15:02
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You dropped a sign in the first line. The $\cos^2 x$ term should be positive inside the parentheses. Now multiply the numerator and denominator by $\cos x$ and you have a sum of cubes in the numerator. The denominator cancels.

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  • $\begingroup$ The signs always get the best of me. $\endgroup$ – AugieJavax98 May 8 '18 at 14:55
  • $\begingroup$ Thanks for the hint $\endgroup$ – AugieJavax98 May 8 '18 at 14:55

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