<s>[INST] 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.
 [/INST] 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.
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