Given that the inverse trigonometric functions take principal values only. Then, the number of real values of $x$ which satisfy ${sin}^{- 1} (\frac{3 x}{5}) + {sin}^{- 1} (\frac{4 x}{5}) = {sin}^{- 1} x$ is equal to :

0

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2

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Solution

The correct option is B $3$
${sin}^{- 1} (\frac{3 x}{5}) + {sin}^{- 1} (\frac{4 x}{5}) = {sin}^{- 1} x$
Taking sine on both sides
$\frac{3 x}{5} \sqrt{1 - \frac{16 x^{2}}{25}} + \frac{4 x}{5} \sqrt{1 - \frac{9 x^{2}}{25}} = x$
$\Rightarrow 3 x \sqrt{25 - 16 x^{2}} = 25 x - 4 x \sqrt{25 - 9 x^{2}}$
$\Rightarrow x = 0$
or $3 \sqrt{25 - 16 x^{2}} = 25 - 4 \sqrt{25 - 9 x^{2}}$
$\Rightarrow 9 (25 - 16 x^{2}) = 625 - 200 \sqrt{25 - 9 x^{2}} + 16 (25 - 9 x^{2})$
$\Rightarrow 200 \sqrt{25 - 9 x^{2}} = 800$
$\Rightarrow \sqrt{25 - 9 x^{2}} = 4$
$\Rightarrow x^{2} = 1$
$\Rightarrow x = \pm 1$
$∴$ Total number of solutions $= 3$

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Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin−1(3x5)+sin−1(4x5)=sin−1x is equal to :

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