Question

Solve the equation for x: `sin^(-1)  5/x + sin^(-1)  12/x = π/2, x ≠ 0`

Solve the equation x: `sin^(-1) (5/x) + sin^(-1) (12/x) = π/2 (x ≠ 0)`

Answer 1

`sin^(-1) (5/x) + sin^(-1) (12/x) = π/2`

`sin^(-1) + cos^(-1) sqrt(1 - 144/x^2) = π/2`

Let `sin^(-1)  12/x = β`

`12/x = sin β = "OPP"/"HYP"`

`sqrt(x^2 - 144)/x = cos β = "Adj"/"HYP"`

`β = cos^(-1) (sqrt(x^2 - 144)/x^2)`

∴ `5/x = sqrt(1 - 144/x^2)`

`25/x^2 = 1 - 144/x^2`

`169/(x^2) = 1`

x² = 169

x = 13

Answer 2

`sin^(-1) (5/x) + sin^(-1) (12/x) = π/2`

⇒ `sin^-1 (5/x) = π/2 - sin^-1 (12/x)`

⇒ `sin^-1 (5/x) = cos^-1 (12/x)`   ...`[∵ sin^-1x + cos^-1x = π/2]`

⇒ `sin^-1 (5/x) = cos^-1 (12/x) = θ`

⇒ `sin θ = 5/x, cos θ = 12/x`

Since, sin²θ + cos²θ = 1

⇒ `(5/x)^2 + (12/x)^2 = 1`

⇒ `25/(x^2) + 144/(x^2) = 1`

⇒ `169/(x^2) = 1`

⇒ x = ±13

Since, x = –13 does not satisfy the given equation.

So, x = 13.