Prove that: sin θ/(cot θ + cosec θ) = 2 + [sin θ/ (cot θ – cosec θ

Sure, let's prove the given equation step by step.

Given equation is: sin θ/(cot θ + cosec θ) = 2 + sin θ/ (cot θ – cosec θ)

Step 1: Convert all trigonometric ratios to sine and cosine.

So, the equation becomes: sin θ/(cos θ/sin θ + 1/sin θ) = 2 + sin θ/ (cos θ/sin θ – 1/sin θ)

Step 2: Simplify the equation.

This gives us: sin^2 θ/(cos θ + 1) = 2 + sin^2 θ/ (cos θ – 1)

Step 3: Cross multiply to get rid of the denominators.

This gives us: sin^2 θ (cos θ - 1) = 2(cos θ + 1) + sin^2 θ (cos θ + 1)

Step 4: Expand and simplify.

This gives us: sin^2 θ cos θ - sin^2 θ = 2 cos θ + 2 + sin^2 θ cos θ + sin^2 θ

Step 5: Cancel out the common terms on both sides.

This leaves us with: - sin^2 θ = 2 cos θ + 2

Step 6: Rearrange the equation.

This gives us: sin^2 θ + 2 cos θ + 2 = 0

Step 7: Use the Pythagorean identity sin^2 θ + cos^2 θ = 1.

This gives us: 1 + 2 cos θ + 2 = 0

Step 8: Simplify the equation.

This gives us: 2 cos θ + 3 = 0

Step 9: Solve for cos θ.

This gives us: cos θ = -3/2

However, the range of cos θ is from -1 to 1. Therefore, there is no real solution for this equation. Hence, the given equation is not valid.