Question

d/dx(sec^2x-tan^2x)=?
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Answer 1

We know that

$1 + \tan( {x})^{2} = \sec( {x})^{2}$

So,

${ \sec(x) }^{2} - { \tan(x) }^{2} = 1$

Diffrentiating both sides, we know that diffrentiation of constant is 0. So,

$dy \div dx = 0$