Question

d/dx 1/tan ( secx + tanx)

Answer 1

Answer :

Let,

y = $\frac{1}{tan(secx+tanx)}$

⇒ y = cot(secx+tanx)

Now, differentiating both sides with respect to x, we get

$\frac{dy}{dx} = \frac{d}{dx} cot(secx + tanx) \\ \\ = - {cosec}^{2} (secx + tanx) \times \frac{d}{dx} (secx + tanx) \\ \\ = - {cosec}^{2} (secx + tanx) \times (secx \: tanx + {sec}^{2} x)$

which is the required derivative

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