Question

Given that the inverse trigonometric function take principal values only. Then, the number of real values of x which satisfy `sin^-1((3x)/5) + sin^-1((4x)/5) = sin^-1x` is equal to ______.

Options

• 3

• 1

• 0

• 2

Answer 1

Given that the inverse trigonometric function take principal values only. Then, the number of real values of x which satisfy `sin^-1((3x)/5) + sin^-1((4x)/5)` = sin^–1x is equal to 3.

Explanation:

`sin^-1((3x)/5) + sin^-1((4x)/5)` = sin^–1x  ...[∵ sin^–1x + sin^–1y = `sin^-1(xsqrt(1 - y^2) + ysqrt(1 - x^2))`]

`\implies sin^-1((3x)/5 sqrt(1 - (16x^2)/25) + (4x)/5 sqrt(1 - (9x^2)/25))` = sin^–1x

`\implies sin^-1((3xsqrt(25 - 16x^2) + 4xsqrt(25 - 9x^2))/25)` = sin^–1x

`\implies 3xsqrt(25 - 16x^2) + 4xsqrt(25 - 9x^2)` = 25x

`\implies` x = 0 or

`\implies` 9(25 – 16x²) = `625 + 16(25 - 9x^2) - 200sqrt(25 - 9x^2)`

`\implies 200sqrt(25 - 9x^2)` = 800

`\implies` x² = 1

`\implies` x = ±1

Total number of solutions = 3