Answer
true
Work Step by Step
$\frac{d}{dx}tan^{2}x=\frac{d}{dx}(sec^{2}x)$
Recall the property: $sec^{2}\theta=1+tan^{2}\theta$
$\frac{d}{dx}tan^{2}x=\frac{d}{dx}(1+tan^{2}x)$
$\frac{d}{dx}tan^{2}x=0+\frac{d}{dx}(tan^{2}x)$
or
$\frac{d}{dx}(tan^{2}x)=\frac{d}{dx}(tan^{2}x)$
Hence, the given statement is true.
