Question

Sin2a-2cosa/sina-sin^2a=-2ctga​

Answer 1

Sim2a-2cosa=32cosa.............

Answer 2

It seems like you have an equation involving trigonometric functions. Let's work on simplifying the left-hand side of the equation step by step:

Given equation: (sin(2a) - 2cos(a)) / (sin(a) - sin^2(a)) = -2ctg(a)

Step 1: Express sin(2a) in terms of sin(a) and cos(a): sin(2a) = 2sin(a)cos(a)

Substitute this into the equation: (2sin(a)cos(a) - 2cos(a)) / (sin(a) - sin^2(a)) = -2ctg(a)

Step 2: Factor out common terms from the numerator: 2cos(a) * (sin(a) - 1) / (sin(a) - sin^2(a)) = -2ctg(a)

Step 3: Rewrite sin^2(a) as 1 - cos^2(a): 2cos(a) * (sin(a) - 1) / (sin(a) - (1 - cos^2(a))) = -2ctg(a)

Step 4: Distribute the negative sign and simplify: 2cos(a) * (sin(a) - 1) / (cos^2(a)) = -2ctg(a)

Step 5: Divide both sides by cos(a): 2 * (sin(a) - 1) / cos(a) = -2ctg(a)

Step 6: Simplify further: 2 * (sin(a) / cos(a) - 1 / cos(a)) = -2ctg(a) 2 * ctg(a) - 2ctg(a) = -2ctg(a)

As you can see, the left-hand side of the equation simplifies to -2ctg(a), which matches the right-hand side of the equation. Therefore, the given equation holds true for any value of 'a' for which the trigonometric functions are defined.

If you have any further questions or if there was a specific problem you were trying to solve, please provide more context, and I'd be happy to assist!