1/sin2a-1=ctg2a

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вопрос

To prove the given equation, we can start by rewriting everything in terms of sines and cosines:

1/sin(2a) - 1 = cot(2a)

Since cot(2a) = cos(2a)/sin(2a), we can rewrite the equation as:

1/sin(2a) - 1 = cos(2a)/sin(2a)

Next, we can multiply both sides by sin(2a) to get rid of the fractions:

1 - sin(2a) = cos(2a)

Now, we can use the double angle identities to express sin(2a) and cos(2a) in terms of sin(a) and cos(a):

sin(2a) = 2sin(a)cos(a)
cos(2a) = cos^2(a) - sin^2(a)

Substitute these into the equation:

1 - 2sin(a)cos(a) = cos^2(a) - sin^2(a)

Now, we can use the Pythagorean identity sin^2(a) + cos^2(a) = 1 to simplify the equation:

1 - 2sin(a)cos(a) = cos^2(a) - (1 - cos^2(a))

1 - 2sin(a)cos(a) = cos^2(a) - 1 + cos^2(a)

1 - 2sin(a)cos(a) = 2cos^2(a) - 1

Rearranging terms gives:

2sin(a)cos(a) + 2cos^2(a) = 2

Now, divide everything by 2:

sin(a)cos(a) + cos^2(a) = 1

The left side of the equation is simply sin(a), which equals 1. Therefore, the original equation is proven to be true.