Let f(x)=sin−1(2x1+x2) and g(x)=cos−1(x2−1x2+1) , then the value of (f(10)−g(100)) is equal to

f(x)=sin−1(2x1+x2)=π−2tan−1x, for x≥1and g(x)=cos−1(x2−1x2+1)=π−cos−1(1−x21+x2) g(x)=π−2tan−1x,for x≥0Now f(10)−g(100)=(π−2tan−1(10))−(π−2tan−1(100))(f(10)−g(100))=2(tan−1(100)−tan−1(10))

Let f(x)=sin−1(2x1+x2) and g(x)=cos−1(x2−1x2+1) , then the value of (f(10)−g(100)) is equal to

f(x)=sin−1(2x1+x2)=π−2tan−1x, for x≥1and g(x)=cos−1(x2−1x2+1)=π−cos−1(1−x21+x2) g(x)=π−2tan−1x,for x≥0Now f(10)−g(100)=(π−2tan−1(10))−(π−2tan−1(100))(f(10)−g(100))=2(tan−1(100)−tan−1(10))

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Let $f (x) = \sin^{- 1} (\frac{2 x}{1 + x^{2}})$ and $g (x) = \cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1})$ , then the value of $(f (10) - g (100))$ is equal to

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$π - 2 (\tan^{- 1} (10) + \tan^{- 1} (100))$

$2 (\tan^{- 1} (10) - \tan^{- 1} (100))$

$2 (\tan^{- 1} (100) - \tan^{- 1} (10))$

answer is C.

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Detailed Solution

$f (x) = \sin^{- 1} (\frac{2 x}{1 + x^{2}}) = π - 2 \tan^{- 1} x, for x \geq 1$
and $g (x) = \cos^{- 1} (\frac{x^{2} - 1}{x^{2} + 1}) = π - \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}})$