Prove that `1+cot theta <= cot (theta/2) `for `0 < theta < pi`. Find `theta` when equality signs holds.

Explanation

`1+cot theta - cot(theta/2) = 1+(cot^2(theta/2) - 1)/(2cot(theta/2)) - cot(theta/2)`

`=(2cot(theta/2)+cot^2(theta/2)-1 -2cot^2(theta/2))/(2cot(theta/2))`

`=-(cot^2(theta/2)-2cot(theta/2)+1)^2/(2cot(theta/2))`

`=-(cot theta-1)^2/(2cot(theta/2))`

As `-(cot (theta/2)-1)^2/(2cot(theta/2))` is always less than or equal to `0`,

`:. 1+cot theta - cot(theta/2) le 0`

`=>1+cot theta le cot(theta/2).`

Now, when equality sign holds,

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