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I have to simplify (the answer is $\sec(x)$):

$$\frac{\sin(x)+\sin(x)\cdot\tan^2(x)}{\tan(x)}$$

I have looked at images for all trig identities but nothing shows $\sin(x)+\sin(x)$ or $\frac{\sin\left(x\right)+\sin\left(x\right)}{\tan\left(x\right)}$

In short: I tried brute tests on the calculator, but different values give differing answers.

Longer explanation: I tried to test random values (but the same value for each function), and keep getting differing results. Such as $\tan(45)^2 = 1$ in degree mode, but $\tan(5)^2 != 1$. Tried in Radian mode and the results are $2.62$ and $11.43$ respectively. So couldn't pick out a pattern.

**EDIT

Thank you for those who have provided answers. I really can not figure out what happened to my original $sin(x) + $ in all the answers provided.

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    $\begingroup$ What do you mean that the answer is $\sin()$ in the parenthesis in the first line? $\endgroup$ – awllower Mar 12 '20 at 2:44
  • $\begingroup$ Sorry it should be sec(x). Thank you for pointing it out. $\endgroup$ – Anon Mar 12 '20 at 3:06
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In these types of questions, the general method is to convert the expression into sines and cosines. Letting $s=\sin x,c=\cos x$, we get $$\begin{split}\frac{\sin(x)+\sin(x)\tan(x)^2}{\tan(x)}&=\frac{s+s(s/c)^2}{s/c}\\&=\frac{sc^2+s^3}{sc}\\&=\frac{s^2+c^2}{c}\\&=\frac1c=\sec(x)\end{split}$$ where we used that $\tan(x)=\sin(x)/\cos(x)=s/c$, and $\sin(x)^2+\cos(x)^2=1$, i.e. $s^2+c^2=1$.

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  • $\begingroup$ Hi @YiFan.. What happened to the sin(x) + at step 2? $\endgroup$ – Anon Mar 13 '20 at 5:24
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    $\begingroup$ @Anon we multiply numerator and denominator by $c^2$, so the $s$ became $sc^2$. $\endgroup$ – YiFan Mar 13 '20 at 5:35
  • $\begingroup$ Sorry to ask, but why do we multiply by $c^2$? Is it because inside the numerator we had $(s/c)^2$, so we multiplied both numerator (which included sin(x)+) and the denominator, which inside the denominator had $s/c x c^2)$ and this would have just.....now I am confused lol I can't find any identities that turn it into just cosine in the denominator. $\endgroup$ – Anon Mar 13 '20 at 5:43
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    $\begingroup$ @Anon no need to rush to turn the denominator into $\cos(x)$. We aim to simplify everything first and hope that the correct answer comes out in the end (and of course, it does). We multiplied by $c^2$ because there was a $(s/c)^2$ in the numerator; nested fractions are annoying, so we simplify by getting rid of this nested fraction, multiplying by $c^2$ to clear the denominator of $(s/c)^2$ to make it just $c^2(s/c)^2=s^2$. $\endgroup$ – YiFan Mar 13 '20 at 6:18
  • $\begingroup$ Finally got it, thank you so much. I was missing "a+ab" = a(1+b)", but with that, and your last comment your's was the most helpful. Thank you so much YiFan for taking the time to help. $\endgroup$ – Anon Mar 13 '20 at 6:38
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Using $$1 + \tan^{2}(x) = \frac{\cos^{2}(x) + \sin^{2}(x)}{\cos^{2}(x)} = \frac{1}{\cos^{2}(x)}$$ then $$\frac{\sin(x) \, (1 + \tan^{2}(x))}{\tan(x)} = \frac{\sin(x)}{\cos^{2}(x) \, \tan(x)} = \frac{1}{\cos(x)}.$$

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  • $\begingroup$ Hi @Leucippus. Why is your second equation missing the "sin(x) +"? What happened to it? thank you $\endgroup$ – Anon Mar 13 '20 at 5:32
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This is equivalent to $$\frac{\sin x}{\tan x} + \sin x \tan x = \cos x + \frac{\sin^2x}{\cos x} = \cos x + \frac{(1-\cos^2x)}{\cos x} = \frac{1}{\cos x}.$$ Or, you can use it is equivalent to $$\frac{\sin x (1+\tan^2x)}{\tan x} = \frac{(\sin x \frac{1}{\cos^2x})}{\tan x} = \frac{\cos x }{\cos^2x} = \frac{1}{\cos x}.$$

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  • $\begingroup$ Sorry that I don't know LaTEX. $\endgroup$ – user690234 Mar 12 '20 at 2:39
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    $\begingroup$ The identities I used were sin^2=1-cos^2(method 1), and 1+tan^2=1/cos^2(method 2). $\endgroup$ – user690234 Mar 12 '20 at 2:41
  • $\begingroup$ Hi @user690234, What happened to the sin(x) +, in your "equivalent" step? $\endgroup$ – Anon Mar 13 '20 at 5:36
  • $\begingroup$ ah yes thankyou for correcting me $\endgroup$ – user690234 Mar 13 '20 at 9:45
  • $\begingroup$ sorry i don't know latex so somebody please help me fix it $\endgroup$ – user690234 Mar 13 '20 at 9:47
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This question is easily solve by factoring and knowing that $1+\tan^2{x}=\sec^2x$

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

In almost any mathematics excersise you should factorize whenever you can, it gives you a wider view of the problem and it can make it easier. Even when computing basic algebra calculations such as division, factorizing can be the key to solving them quickly and flawless

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