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How do you identify \[\dfrac{1}{\cot^{2}x + 1}\] ?

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Last updated date: 22nd Jul 2024
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Answer
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Hint: First , we can use the trigonometric identity \[\cot^{2}x + 1 =\text{cosec}^{2}x\] . Then , we need to rewrite the Cosecant function in terms of the sine function. Using this, we can evaluate the given expression.
Identity used :
\[\cot^{2}x + 1 =\text{cosec}^{2}x\]
\[\text{cosec}\ x = \dfrac{1}{\sin x}\]

Complete step by step answer:
Given,
\[\dfrac{1}{\cot^{2}x + 1}\]
By using the identity \[\cot^{2}x + 1 = \text{cosec}^{2}x\],
We get,
\[\dfrac{1}{\cot^{2}x + 1} = \dfrac{1}{\text{cosec}^{2}x}\]
We can rewrite the Cosecant function in terms of the sine function.
We know that
\[\text{cosec}\ x = \dfrac{1}{\sin x}\]
Thus we get,
\[\Rightarrow \dfrac{1}{\left( \dfrac{1}{\sin^{2}x} \right)}\]
On taking reciprocal,
We get,
\[\Rightarrow \ \sin^{2}x\]
Thus we get \[\dfrac{1}{\cot^{2}x + 1}\] is equal to \[\sin^{2}x\] .

Note:
Students should note that by looking at the given expression we can see that this expression will be simplified directly using identity. We have similar kinds of identities if our expression is in terms of sin & cos and tan & sec as well.