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Simplify $1 + \tan^2x$

My attempt:

$$\begin{align}1 + \tan^2x&\\ &= \frac{1}{1} + \frac{\sin^2x}{\cos^2x}\\ &= \frac{1(\cos^2x)}{1(\cos^2x)} +\frac{\sin^2x}{\cos^2x}\\ &=\frac{\cos^2x}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\ &=\frac{\cos^2x + \sin^2x}{\cos^2x\cos^2x}\\ &= \frac{\sin^2x}{\cos^2x}\\ &= \tan^2x\end{align}$$

The correct answer, however..is $sec^2x$ Wherever I went wrong, please show.

Thomas Andrews
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Qwerty
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    Your fourth line has $\cos^2(x)\cos^2(x)$ as a denominator. I believe it should just be $\cos^2(x)$ – Jam Aug 09 '14 at 14:22
  • Also, you can use the identity $\sin^2(x)+\cos^2(x)=1$. – Jam Aug 09 '14 at 14:24
  • Yes, if $$\frac{\cos^2 x}{\cos^2 x} + \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2+\sin^2 x}{\cos^2x\cos^2x} = \frac{\sin^2 x}{\cos^2 x}$$ then $\frac{\cos^2 x}{\cos^2 x} = 0$. (In fact, both equalities are wrong.) – Thomas Andrews Aug 09 '14 at 14:35

4 Answers4

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$${\cos^2x\over\color{red}{\cos^2x}}+{\sin^2x\over\color{red}{\cos^2x}}={\cos^2x+\sin^2x\over\color{red}{\cos^2x}}={1\over\cos^2x}=\sec^2x.$$

Jaehyeon Seo
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    Spent 40mins re-doing this, I made the same mistake I have been doing since 4th grade, 7 years ago... Thank you for the insight. My memory sucks :( – Qwerty Aug 09 '14 at 14:23
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$$1 + \tan^2 x \implies \sec^2 x - \tan^2 x + \tan^2 x \implies \sec^2 x$$

ki3i
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Anunoy Chakraborty
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Firstly, I think that everyone (including yourself) disagrees with the assertion that in general $$\frac{a}{a}+\frac{b}{a}=\frac{a+b}{a^2}$$ by the distributive law $$\frac{a}{a}+\frac{b}{a}=\frac{1}{a}(a+b)=\frac{a+b}{a}$$ Secondly, I think that $\tan^2x+1=\sec^2x$ should be considered to be just as basic of an identity as $\sin^2x+\cos^2x=1$ is. My reasoning is as follows. Taking the Pythagorean relationship $$opposite^2+adjacent^2=hypotenuse^2$$ and dividing both sides by $hypotenuse^2$ (yielding $\sin^2 x+\cos^2x=1$) is just as simple as dividing said relationship by $adjacent^2$, yielding ($\tan^2x+1=\sec^2x$). Because of that reasoning, I would either just state the identity $$\tan^2x+1=\sec^2x$$ or perform a more elemental proof $$\begin{array}{lll} opposite^2+adjacent^2&=&hypotenuse^2\\ \frac{opposite^2+adjacent^2}{adjacent^2}&=&\frac{hypotenuse^2}{adjacent^2}\\ \frac{opposite^2}{adjacent^2}+\frac{adjacent^2}{adjacent^2}&=&\frac{hypotenuse^2}{adjacent^2}\\ \bigg(\frac{opposite}{adjacent}\bigg)^2+\bigg(\frac{adjacent}{adjacent}\bigg)^2&=&\bigg(\frac{hypotenuse}{adjacent}\bigg)^2\\ \tan^2x+1&=&\sec^2x& \end{array}$$ unless instructed to do otherwise

John Joy
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4 th line of your solution is clearly wrong. In denominator their must be lcm of [(cosx)^2 * (cosx)^2] which is (cosx)^2 and in 5th line you dont have used well known identity (cosx)^2 + (sinx)^2 = 1.

Akash Patalwanshi
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