If f(x)=cot−1(2x1−x2) and g(x)=cos−1(1−x21+x2) , then limx→af(x)−f(a)g(x)−g(a)(0<a<12) is

f(x)−f(a)g(x)−g(a)={π2−tan−1(2x1−x2)}−{π2−tan−1(2a1−a2)}cos−1(1−x21+x2)−cos−1(1−a21+a2) =tan−1(2a1−a2)−tan−1(2x1−x2)cos−1(1−x21+x2)−cos−1(1−a21+a2)=tan−1(tan2α)−tan−1(tan2θ)cos−1(cos2θ)−cos−1(cos2α) , where x=tanθ,a=tanα . =2(α−θ)2(θ−α)=−1

If f(x)=cot−1(2x1−x2) and g(x)=cos−1(1−x21+x2) , then limx→af(x)−f(a)g(x)−g(a)(0<a<12) is

f(x)−f(a)g(x)−g(a)={π2−tan−1(2x1−x2)}−{π2−tan−1(2a1−a2)}cos−1(1−x21+x2)−cos−1(1−a21+a2) =tan−1(2a1−a2)−tan−1(2x1−x2)cos−1(1−x21+x2)−cos−1(1−a21+a2)=tan−1(tan2α)−tan−1(tan2θ)cos−1(cos2θ)−cos−1(cos2α) , where x=tanθ,a=tanα . =2(α−θ)2(θ−α)=−1

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If $f (x) = \cot^{- 1} (\frac{2 x}{1 - x^{2}})$ and $g (x) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$ , then $\lim_{x \to a} \frac{f (x) - f (a)}{g (x) - g (a)} (0 < a < \frac{1}{2})$ is

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Detailed Solution

$\frac{f (x) - f (a)}{g (x) - g (a)} = \frac{{\frac{π}{2} - \tan^{- 1} (\frac{2 x}{1 - x^{2}})} - {\frac{π}{2} - \tan^{- 1} (\frac{2 a}{1 - a^{2}})}}{\cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) - \cos^{- 1} (\frac{1 - a^{2}}{1 + a^{2}})}$

$= \frac{\tan^{- 1} (\frac{2 a}{1 - a^{2}}) - \tan^{- 1} (\frac{2 x}{1 - x^{2}})}{\cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) - \cos^{- 1} (\frac{1 - a^{2}}{1 + a^{2}})} = \frac{\tan^{- 1} (\tan 2 α) - \tan^{- 1} (\tan 2 θ)}{\cos^{- 1} (\cos 2 θ) - \cos^{- 1} (\cos 2 α)}$ , where $x = \tan θ, a = \tan α$ .