Answer:
x = π/2 or
x = π/2 + 2πn (for any value of n)
Explanation:
We need to find the numerical value of the trigonometric function sinx+1/sinx=1+cscx
Subtract 1+cscx on both sides
sinx + 1/sinx -(1+csc x) = 1+cscx - (1+cscx)
sinx + 1/sinx -1 -csc x = 0
Taking LCM i.e, sinx and solving
(sinx)(sinx) + 1 - sinx -cscx(sinx) / sinx = 0
Multiplying sinx on both sides
sin^2 x + 1 -sinx - cscx(sinx) =0
as we know, cscx = 1/ sinx
sin^2 x + 1 -sinx - (1/sinx)(sinx) = 0
Solving,
sin^2 x + 1 -sinx - 1 = 0
sin^2 x - sin x =0
Let u = sinx
Putting it in above equation
u^2 - u =0
Solving the equation:
u= 1 , u= 0
Putting back the value of u
sin(x) = 1 and sin(x) = 0
x = sin ⁻¹ (1) and x = sin ⁻¹ (1) =0
x = 90° or π/2 and x = 0 (undefined for the question)
So, x = π/2 or
x = π/2 + 2πn (for any value of n)