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UPSC NDA Math Model Paper 1

Question : 9 of 120

Marks: +1, -0

f(x)=\tan^{-1}\left[\frac{\sin x}{1+\cos x}\right],

f(x)?

Solution:

\;\text{ Concept: }\;

Derivatives of Trigonometric Functions: $\frac{d}{dx} \sin x = \cos x \quad \frac{d}{dx} \cos x$ $=-\sin x \;\frac{d}{dx} \tan x = \sec^2 x$
$\frac{d}{dx} \cot x = -\csc^2 x \;\frac{d}{dx} \sec x$ $=\tan x \sec \frac{d}{dx} \csc x = -\cot x \csc$
Trigonometric Identities: $\sin^2 \theta + \cos^2 \theta$ $=1$ $\sin 2\theta$ $=2 \sin \theta \cos \theta$ $\cos 2\theta$ $=\cos^2 \theta - \sin^2 \theta$

Calculation:

Let us express

\;\frac{\sin x}{1+\cos x}

in terms of tan

x

\;\frac{\sin x}{1+\cos x}

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

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

\therefore f(x)=\tan^{-1}\left[\frac{\sin x}{1+\cos x}\right]

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

And, the first derivative of

f(x)=f'(x)=\frac{d}{dx}\left(\frac{x}{2}\right)=\frac{1}{2}

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