MCQ
If $AB = A$ and $BA = B$, then
- A$A^2B = A^2$
- B$B^2A = B^2$
- C$ABA = A$
- ✓All of the above
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General solution of $\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=\sec\text{x}$ is:
$\text{y}\sec\text{x}=\tan\text{x}+\text{c}$
$\text{y}\tan\text{x}=\sec\text{x}+\text{c}$
$\tan\text{x}=\sec\text{x}+\text{c}$
$\text{x}\sec\text{x}=\tan\text{y}+\text{c}$