Multiple choice

If $\sin x+ \mathrm{cosec} x+ \tan y+ \cot y=4$, where x and y $\epsilon [0, \dfrac {\pi}{2}]$, then $\tan\dfrac {y}{2}$ is a root of the equation-

  1. $\alpha^2+2\alpha+1=0$
  2. $\alpha^2+2\alpha-1=0$
  3. $2\alpha^2-2\alpha-1=0$
  4. $\alpha^2-\alpha -1=0$
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B Correct answer
Explanation

Using the AM-GM inequality, sin x + cosec x >= 2 and tan y + cot y >= 2. Since the sum is 4, both must equal 2, implying sin x = 1 and tan y = 1. If tan y = 1, then y = pi/4, so tan(y/2) = tan(pi/8) = sqrt(2) - 1. This value satisfies the equation alpha^2 + 2*alpha - 1 = 0.

AI explanation

For angles in the interval [0, pi/2], both sin x + cosec x and tan y + cot y have a minimum value of 2 by the AM-GM inequality, meaning the sum equals 4 only when sin x = 1 and tan y = 1. This gives y = pi/4, so we evaluate tan(y/2) as tan(pi/8), which equals sqrt(2) - 1. Substituting alpha = sqrt(2) - 1 into the identity alpha^2 + 2alpha - 1 = (sqrt(2) - 1)^2 + 2(sqrt(2) - 1) - 1 gives 0, proving it is the root of alpha^2 + 2alpha - 1 = 0.

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