If cot(cos-1 3/5 + sin-1 x) = 0, find the value of x.

1.	If cot(cos-1 3/5 + sin-1 x) = 0, find the value of x.
Answer» Given as cot(cos^-13/5 + sin^-1 x) = 0 By, rearranging (cos^-13/5 + sin^-1 x) = cot^-1 (0) (cos^-13/5 + sin^-1 x) = π/2 ...(i) As we know that cos^-1x + sin^-1 x = π/2 Then sin^-1 x = π/2 – cos^-1x Substitute the above in eq. (i) we get, (cos^-13/5 + π/2 – cos^-1x) = π/2 By, rearranging we get, (cos^-13/5 – cos^-1x) = π/2 – π/2 (cos^-13/5 – cos^-1x) = 0 So, cos^-13/5 = cos^-1x By, comparing the above equation we get, x = 3/5

Answer»

Given as cot(cos^-13/5 + sin^-1 x) = 0

By, rearranging

(cos^-13/5 + sin^-1 x) = cot^-1 (0)

(cos^-13/5 + sin^-1 x) = π/2 ...(i)

As we know that cos^-1x + sin^-1 x = π/2

Then sin^-1 x = π/2 – cos^-1x

Substitute the above in eq. (i) we get,

(cos^-13/5 + π/2 – cos^-1x) = π/2

By, rearranging we get,

(cos^-13/5 – cos^-1x) = π/2 – π/2

(cos^-13/5 – cos^-1x) = 0

So, cos^-13/5 = cos^-1x

By, comparing the above equation we get,

x = 3/5

Discussion