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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, 2020 at 2:44
  • $\begingroup$ Sorry it should be sec(x). Thank you for pointing it out. $\endgroup$
    – Anon
    Mar 12, 2020 at 3:06

4 Answers 4

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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, 2020 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, 2020 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, 2020 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, 2020 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, 2020 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, 2020 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, 2020 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, 2020 at 2:41
  • $\begingroup$ Hi @user690234, What happened to the sin(x) +, in your "equivalent" step? $\endgroup$
    – Anon
    Mar 13, 2020 at 5:36
  • $\begingroup$ ah yes thankyou for correcting me $\endgroup$
    – user690234
    Mar 13, 2020 at 9:45
  • $\begingroup$ sorry i don't know latex so somebody please help me fix it $\endgroup$
    – user690234
    Mar 13, 2020 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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