Question:
Given that the inverse trigonometric functions take principal values only. Then, the number of real values of \(x\) which satisfy
\(\sin^{-1}\left(\frac{3x}{5}\right) + \sin^{-1}\left(\frac{4x}{5}\right) = \sin^{-1}x\)
is equal to :
Given that the inverse trigonometric functions take principal values only. Then, the number of real values of \(x\) which satisfy
\(\sin^{-1}\left(\frac{3x}{5}\right) + \sin^{-1}\left(\frac{4x}{5}\right) = \sin^{-1}x\)
is equal to :
\(\sin^{-1}\left(\frac{3x}{5}\right) + \sin^{-1}\left(\frac{4x}{5}\right) = \sin^{-1}x\)
is equal to :
Show Hint
The domain is \(|x|\le1\); take sine of both sides, solve, then verify.
Updated On: Nov 4, 2026
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The Correct Option is C
Solution and Explanation
For \(\sin^{-1}x\) to be defined, \(|x|\le1\) (then \(\left|\frac{3x}{5}\right|,\left|\frac{4x}{5}\right|\le1\) automatically).
Take sine of both sides: \(\frac{3x}{5}\sqrt{1 - \frac{16x^2}{25}} + \frac{4x}{5}\sqrt{1 - \frac{9x^2}{25}} = x\).
\(x = 0\) is a solution. For \(x\ne0\): \(3\sqrt{25 - 16x^2} + 4\sqrt{25 - 9x^2} = 25\).
The left side is strictly decreasing in \(x^2\); at \(x^2 = 1\) it equals \(3\cdot3 + 4\cdot4 = 25\), so \(x^2 = 1\) is the only solution, i.e. \(x = \pm1\).
Verify: \(x = 1\): \(\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{4}{5} = \frac{\pi}{2} = \sin^{-1}1\), true; \(x = -1\) is true by oddness.
Solutions: \(x = -1, 0, 1\), i.e. 3 values.
Answer: (3) 3
Take sine of both sides: \(\frac{3x}{5}\sqrt{1 - \frac{16x^2}{25}} + \frac{4x}{5}\sqrt{1 - \frac{9x^2}{25}} = x\).
\(x = 0\) is a solution. For \(x\ne0\): \(3\sqrt{25 - 16x^2} + 4\sqrt{25 - 9x^2} = 25\).
The left side is strictly decreasing in \(x^2\); at \(x^2 = 1\) it equals \(3\cdot3 + 4\cdot4 = 25\), so \(x^2 = 1\) is the only solution, i.e. \(x = \pm1\).
Verify: \(x = 1\): \(\sin^{-1}\frac{3}{5} + \sin^{-1}\frac{4}{5} = \frac{\pi}{2} = \sin^{-1}1\), true; \(x = -1\) is true by oddness.
Solutions: \(x = -1, 0, 1\), i.e. 3 values.
Answer: (3) 3
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