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The least value of expression cot2x-tan2x/1 sin(5π 2)?
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The least value of expression cot2x-tan2x/1 sin(5π 2)?
Calculating the Least Value of the Expression



  • Firstly, we need to simplify the expression cot2x - tan2x/1 sin(5π/2).

  • We know that sin(5π/2) equals -1, so substituting this value, we get cot2x - tan2x/-1.

  • Using the identity cot2x = 1/tan2x, we can simplify the expression further to (1/tan2x) - tan2x/-1.

  • Combining the fractions, we get (1 - tan4x)/-tan2x.

  • To find the least value of the expression, we need to find the maximum value of the denominator and the minimum value of the numerator.

  • The maximum value of the denominator is 0, which occurs when tan2x is undefined or infinite. This happens when x = nπ/2, where n is an integer.

  • The minimum value of the numerator is -1, which occurs when tan4x = 1.

  • Using the identity tan2x = (2tanx)/(1-tan2x), we can solve for tanx and get tanx = ±sqrt((2-sqrt(3))/2) or ±sqrt((2+sqrt(3))/2).

  • Since we want the least value of the expression, we take the negative square root and substitute it into the expression to get (-1 - (-sqrt(3)))/-tan2x = sqrt(3)-1.

  • Therefore, the least value of the expression is sqrt(3)-1.

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The least value of expression cot2x-tan2x/1 sin(5π 2)?
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