<s>[INST] Verifying $\sec^2x + \tan^2x = (1-\sin^4x)\sec^4x$ 
Verify: $$\sec^2x + \tan^2x = (1-\sin^4x)\sec^4x$$

My solution:
$$
\begin{align}\sec^2x+\tan^2x&=\frac{1}{\cos^2x}+\frac{\sin^2x}{\cos^2x}\\
&=\frac{1+\sin^2x}{\cos^2x}\\
&=\frac{1+\sin^2x}{\cos^2x}\cdot\frac{1-\sin^2x}{1-\sin^2x}\\
&=\frac{1-\sin^4x}{\cos^2x\cdot\cos^2x}\\
&=\frac{1-\sin^4x}{\cos^4x}\\
&=\frac{1}{\cos^4x}-\frac{\sin^4x}{\cos^4x}\\
&=\sec^4x-\sin^4x\sec^4x\\
&=\sec^4x(1-\sin^4x)\\
\end{align}$$
Is it incorrect to multiply in $1-\sin^2x$ like in the fourth equality?
 [/INST] It's okay because $1-\sin^2 x=\cos^2 x$ so it can only become $0$ if $\cos x$ is zero, wich can be excluded due to the domain of the equality, thus no division by zero can occur and you may proceed.
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