Question

CosB/1-sinB-cosB/1+sinB=2tgB

Answer 1

Объяснение:

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Answer 2

To simplify the given expression, let's manipulate both sides of the equation separately and then compare them.

Starting with the left side: CosB/(1-sinB) - CosB/(1+sinB)

We can find the common denominator for both terms: [(CosB * (1 + sinB)) - (CosB * (1 - sinB))] / [(1 - sinB) * (1 + sinB)]

Expanding the terms in the numerator: [CosB + CosB * sinB - CosB - CosB * sinB] / [(1 - sinB) * (1 + sinB)]

The terms with sinB cancel each other out, leaving us with: 0 / [(1 - sinB) * (1 + sinB)]

Since the numerator is zero, the entire expression becomes zero.

Now let's simplify the right side of the equation: 2 * tanB

Using the trigonometric identity tanB = sinB / cosB: 2 * (sinB / cosB)

Multiplying through by cosB: (2 * sinB) / cosB

The resulting expression is (2 * sinB) / cosB.

Therefore, we have simplified the left side to 0 and the right side to (2 * sinB) / cosB. Since 0 is not equal to (2 * sinB) / cosB, the given equation CosB/(1-sinB) - CosB/(1+sinB) = 2tanB does not hold true for all values of B.