Given that x is a positive real, the maximum possible value of sin(tan−1(x25)−tan−1(x36)) is equal to ____

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Given that x is a positive real, the maximum possible value of $\sin (\tan^{- 1} (\frac{x}{25}) - \tan^{- 1} (\frac{x}{36}))$ is equal to _____

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$\frac{11}{25}$

$\frac{11}{61}$

$\frac{60}{61}$

$\frac{25}{36}$

answer is B.

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Detailed Solution

Consider a right triangle AOC such that AO = 36 and CO = x
Consider B on AO such that BO = 25 and AB = 11
$t a n B = \frac{x}{25}, t a n A = \frac{x}{36}$
Angle B – Angle A = Angle BCA
Angle BCA is maximum if the circumcircle of BCA is as small as possible and OC becomes a tangent to the circle BCA
$O C^{2}$ = OB.OA = 25.36 = 900
OC = 30
$\sin (\tan^{- 1} (\frac{30}{25}) - \tan^{- 1} (\frac{30}{36})) = \sin (\tan^{- 1} (\frac{6}{5}) - \tan^{- 1} (\frac{5}{6})) = \sin (\tan^{- 1} (\frac{11}{60})) = \frac{11}{61}$