Find the derivative: frac{d}{dx} tan^{-1}(sec | vedclass

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Question

(B) Let $y = 	an^{-1}(\sec x + 	an x)$.We can rewrite the expression inside the inverse tangent function as:$\sec x + 	an x = \frac{1}{\cos x} + \f

Vedclass · Accepted Answer

$1/2$

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(B) Let $y = 	an^{-1}(\sec x + 	an x)$.We can rewrite the expression inside the inverse tangent function as:$\sec x + 	an x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x}$.Using trigonometric identities $\sin x = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$,$\cos x = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2})$,and $1 = \sin^2(\frac{x}{2}) + \cos^2(\frac{x}{2})$:$\frac{1 + \sin x}{\cos x} = \frac{\cos^2(\frac{x}{2}) + \sin^2(\frac{x}{2}) + 2\sin(\frac{x}{2})\cos(\frac{x}{2})}{\cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2})} = \frac{(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))^2}{(\cos(\frac{x}{2}) - \sin(\frac{x}{2}))(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))} = \frac{\cos(\frac{x}{2}) + \sin(\frac{x}{2})}{\cos(\frac{x}{2}) - \sin(\frac{x}{2})}$.Dividing numerator and denominator by $\cos(\frac{x}{2})$,we get:$\frac{1 + 	an(\frac{x}{2})}{1 - 	an(\frac{x}{2})} = 	an(\frac{\pi}{4} + \frac{x}{2})$.Thus,$y = 	an^{-1}(	an(\frac{\pi}{4} + \frac{x}{2})) = \frac{\pi}{4} + \frac{x}{2}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{d}{dx}(\frac{\pi}{4} + \frac{x}{2}) = 0 + \frac{1}{2} = \frac{1}{2}$.

Find the derivative: $\frac{d}{dx} \tan^{-1}(\sec x + \tan x) = $

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