Given sinx+1/sinx=1+cscx, find a numerical value of one trigonometric function of x.

x = π/2 or

x = π/2 + 2πn (for any value of n)

Step-by-step 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)

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step-by-step explanation: