Simplifying a trigonometric identity 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.
 A: $${\cos^2x\over\color{red}{\cos^2x}}+{\sin^2x\over\color{red}{\cos^2x}}={\cos^2x+\sin^2x\over\color{red}{\cos^2x}}={1\over\cos^2x}=\sec^2x.$$
A: $$1 + \tan^2 x \implies \sec^2 x - \tan^2 x + \tan^2 x \implies \sec^2 x$$
A: 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
A: 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.
