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Army Group X Practice Set 2

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Question : 62 of 70

Marks: +1, -0

rac{d}{dx}\{ an^{-1}(\sec x- an x)\}=?

$-\frac{1}{2}$
1
- 1
$\frac{1}{2}$

Solution:

rac{d}{dx}\{ an^{-1}(\sec x- an x)\}

=\frac{d}{dx}\{\tan^{-1}(\frac{1}{\cos x}-\frac{\sin x}{\cos x})\}

= \frac{d}{dx} \{ \tan^{-1} ( \frac{1 - \sin x}{\cos x} ) \} = \frac{d}{dx}

\tan^{-1}\left(\frac{\sin^{2}\frac{x}{2}+\cos^{2}\frac{x}{2}-2\sin\frac{x}{2}\cos\frac{x}{2}}{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}\right)

\because \sin^{2}x+\cos^{2}x=1

and

\sin x=2\sin\frac{x}{2}\cos\frac{x}{2}

and

\cos A = \cos^{2} \frac{A}{2} - \sin^{2} \frac{A}{2}

=\frac{d}{dx}\{\tan^{-1}\left(\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})}\right)\}

[ \because a^{2}-b^{2}=(a+b)(a-b)

\;\;

and

(a-b)^{2}=a^{2}+b^{2}-2ab]

=\frac{d}{dx}\{\tan^{-1}\left(\frac{\cos\frac{x}{2}-\sin\frac{x}{2}}{\cos\frac{x}{2}+\sin\frac{x}{2}}\right)\}

= \frac{d}{dx}\{ \tan^{-1}\left( \frac{1-\tan\frac{x}{2}}{1+\tan\frac{x}{2}} \right) \}

[\text{because divided by } \cos\frac{x}{2} \text{ in denominator } ; \text{ and numerator }].

= \frac{d}{dx}\left\{ \tan^{-1}\left( \frac{ \tan\frac{\pi}{4} - \tan\frac{x}{2} }{ 1 + \tan\frac{\pi}{4} \cdot \tan\frac{x}{2} } \right) \right\}

= \frac{d}{dx}\{\tan^{-1}\tan(\frac{\pi}{4}-\frac{x}{2})\} = \frac{d}{dx}(\frac{\pi}{4}-\frac{x}{2})

=-\frac{1}{2}

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