
How do you simplify ${{\cot }^{2}}x+1$ ?
Answer
555.9k+ views
Hint: In the given simplifying question, we can do it with a few simple steps by changing the cot or cotangent in the form of sine and cosine functions. After putting the required value, you need to simplify and you will get the required result. So, let’s see what is the approach to the required question.
Complete Step by Step Solution:
The given question which we need to simplify is ${{\cot }^{2}}x+1$.
The first thing we need to do is to convert the cot or cotangent in the form of sine and cosine function, that means, we get,
$\Rightarrow \cot x=\dfrac{\cos x}{\sin x}$
Now, put the required value of cot in the question, that means, we get,
$={{\left( \dfrac{\cos x}{\sin x} \right)}^{2}}+1$
When we simplify it, we get,
$=\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}+1$
$=\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\sin }^{2}}x}$
Now, we know that, ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ when we apply, we get,
$=\dfrac{1}{{{\sin }^{2}}x}$
Now, we know that, $\dfrac{1}{{{\sin }^{2}}x}$ is the reciprocal of ${{\csc }^{2}}x$ when we apply, we get,
$={{\csc }^{2}}x$
Therefore, after simplifying the question we get ${{\csc }^{2}}x$.
Note:
There is an alternative approach to the above question ${{\cot }^{2}}x+1$.
We just need to replace ${{\cot }^{2}}x+1$with ${{\csc }^{2}}x$, which is the simple formula in trigonometric identities. Now just we need to put the required value in the question, that means, we get,
$={{\csc }^{2}}x$
Therefore, after simplifying the question we get ${{\csc }^{2}}x$
Complete Step by Step Solution:
The given question which we need to simplify is ${{\cot }^{2}}x+1$.
The first thing we need to do is to convert the cot or cotangent in the form of sine and cosine function, that means, we get,
$\Rightarrow \cot x=\dfrac{\cos x}{\sin x}$
Now, put the required value of cot in the question, that means, we get,
$={{\left( \dfrac{\cos x}{\sin x} \right)}^{2}}+1$
When we simplify it, we get,
$=\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}+1$
$=\dfrac{{{\cos }^{2}}x+{{\sin }^{2}}x}{{{\sin }^{2}}x}$
Now, we know that, ${{\cos }^{2}}x+{{\sin }^{2}}x=1$ when we apply, we get,
$=\dfrac{1}{{{\sin }^{2}}x}$
Now, we know that, $\dfrac{1}{{{\sin }^{2}}x}$ is the reciprocal of ${{\csc }^{2}}x$ when we apply, we get,
$={{\csc }^{2}}x$
Therefore, after simplifying the question we get ${{\csc }^{2}}x$.
Note:
There is an alternative approach to the above question ${{\cot }^{2}}x+1$.
We just need to replace ${{\cot }^{2}}x+1$with ${{\csc }^{2}}x$, which is the simple formula in trigonometric identities. Now just we need to put the required value in the question, that means, we get,
$={{\csc }^{2}}x$
Therefore, after simplifying the question we get ${{\csc }^{2}}x$
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