Math Doubts

$\cot{(90^°)}$ value

The cot value when angle of a right triangle equals to $90^°$ is called cot of angle $90$ degrees and it is written as $\cot{(90^°)}$ mathematically in sexagesimal system.

$\cot{(90^°)} \,=\, 0$

The exact value of $\cot{(90^°)}$ is zero mathematically.

Alternative form

The $\cot{(90^°)}$ is written in different ways in alternative form. In other words, it is written as $\cot{\Big(\dfrac{\pi}{2}\Big)}$ in circular system and also written as $\cot{(100^g)}$ in centesimal system.

$(1) \,\,\,$ $\cot{\Big(\dfrac{\pi}{2}\Big)} \,=\, 0$

$(2) \,\,\,$ $\cot{(100^g)} \,=\, 0$

Proof

You learnt that the value of $\cot{\Big(\dfrac{\pi}{2}\Big)}$ is zero exactly. Now, it is time to learn how the value of $\cot{(100^g)}$ is zero exactly in trigonometry.

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