Q52 / 150 · MHT CET 2023 (13 May Shift 1) · Mathematics

Differentiation of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ w.r.t. $\cos ^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}\right)$, is

  • A
    $\frac{1}{2}$
  • Correct answer
    B
    1
  • C
    2
  • D
    $\frac{1}{4}$

Solution

Let $\mathrm{u}=\tan ^{-1}\left[\frac{\sqrt{1+x^2}-1}{x}\right]$ and $v=\cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+x^2}}{2 \sqrt{1+x^2}}}\right]$ Put $x=\tan \theta$, then $\theta=\tan ^{-1} x$ $$ \begin{aligned} \therefore \mathrm{u} & =\tan ^{-1}\left[\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right]-\tan ^{-1}\left[\frac{\sec \theta-1}{\tan \theta}\right] \\ & =\tan ^{-1}\left[\frac{\frac{1}{\cos \theta}-1}{\frac{\sin \theta}{\cos \theta}}\right]=\tan ^{-1}\left[\frac{1-\cos \theta}{\sin \theta}\right] \\ & =\tan ^{-1}\left[\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right]-\tan ^{-1}\left(\tan \frac{\theta}{2}\right)=\frac{\theta}{2} \end{aligned} $$ $$ \therefore \quad \mathrm{u}=\frac{\tan ^{-1} x}{2} $$ $$ \begin{aligned} v & =\cos ^{-1}\left[\sqrt{\frac{1+\sqrt{1+\tan ^2 \theta}}{2 \sqrt{1+\tan ^2 \theta}}}\right] \\ & =\cos ^{-1}\left[\sqrt{\frac{1+\sec \theta}{2 \sec \theta}}\right] \\ & =\cos ^{-1}\left[\sqrt{\frac{1+\frac{1}{\cos \theta}}{\frac{2}{\cos \theta}}}\right] \\ & =\cos ^{-1}\left[\sqrt{\frac{1+\cos \theta}{2}}\right] \\ & =\cos ^{-1}\left(\sqrt{\frac{2 \cos ^2\left(\frac{\theta}{2}\right)}{2}}\right) \\ & =\cos ^{-1}\left(\cos \frac{\theta}{2}\right)-\frac{0}{2} \end{aligned} $$ $$ \therefore \quad v=\frac{\tan ^{-1} x}{2} $$ From (i) and (ii), we get $\mathrm{u}=\mathrm{v}$ $$ \therefore \quad \frac{d u}{d v}=1 $$

Hence, the correct answer is option: 'B'.

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