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# How to simplify $\dfrac{{\csc \left( { - x} \right)}}{{\cot \left( { - x} \right)}}$ ?

Last updated date: 04th Aug 2024
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Hint: csc stands for cosec. $\cos ec\left( x \right)$ is a trigonometric function and is reciprocal of $\sin \left( x \right)$ . $\cos ec\left( x \right)$ can be defined as the ratio between hypotenuse and perpendicular to theta ( $\theta$ ). $\cot \left( x \right)$ is also a trigonometric function and is reciprocal to $\tan \left( x \right)$ . $\cot \left( x \right)$ can be defined as the ratio between base and perpendicular to theta ( $\theta$ ). Both of these trigonometric functions are odd functions.

As we already know that both $\cos ec\left( x \right)$ and $\cot \left( x \right)$ are odd functions, which means that $f\left( { - x} \right) = - f\left( x \right)$ . So, we can say that,
$\Rightarrow \csc \left( { - x} \right) = - \csc \left( x \right)$ ---(1)
$\Rightarrow \cot \left( { - x} \right) = - \cot \left( x \right)$ ---(2)
We are given that $\dfrac{{\csc \left( { - x} \right)}}{{\cot \left( { - x} \right)}}$ ,
Using equation (1) and (2) in given question we get,
$\Rightarrow \dfrac{{\csc \left( x \right)}}{{\cot \left( x \right)}}$ ---(3)
We know that $\cos ec\left( x \right)$ is reciprocal of $\sin \left( x \right)$ which means
$\Rightarrow \csc \left( x \right) = \dfrac{1}{{\sin \left( x \right)}}$ ---(4)
And $\cot \left( x \right)$ is the reciprocal of $\tan \left( x \right)$ which means
$\Rightarrow \cot \left( x \right) = \dfrac{{\cos \left( x \right)}}{{\sin \left( x \right)}}$ ---(5)
Now, substituting the values of equation (4) and (5) in equation (3), we get,
$\Rightarrow \dfrac{{\dfrac{1}{{\sin \left( x \right)}}}}{{\dfrac{{\cos \left( x \right)}}{{\sin \left( x \right)}}}} \\ \Rightarrow \dfrac{1}{{\sin \left( x \right)}} \times \dfrac{{\sin \left( x \right)}}{{\cos \left( x \right)}} \;$
$\sin \left( x \right)$ cuts and cancels off and we get,
$\Rightarrow \dfrac{1}{{\cos \left( x \right)}}$ ---(6)
We know that, $\sec \left( x \right)$ is the reciprocal of $\cos \left( x \right)$ which means that, $\sec \left( x \right) = \dfrac{1}{{\cos \left( x \right)}}$ . Substituting the same in equation (6) and we get,
$\Rightarrow \sec \left( x \right)$
Thus, $\dfrac{{\csc \left( { - x} \right)}}{{\cot \left( { - x} \right)}} = \sec \left( x \right)$ .
So, the correct answer is “Sec x”.

Note: Before solving any trigonometric question, students should keep all the necessary trigonometric identities in their mind. This will help them to solve any question easily. Students should also remember that $\cos ec\left( x \right)$ is not the inverse of $\sin \left( x \right)$ and cannot be written as ${\sin ^{ - 1}}x$ . The same applies to $\sec \left( x \right)$ as well as $\cot \left( x \right)$ . Students should keep in mind that $\sec \left( x \right)$ is the reciprocal of $\cos \left( x \right)$ , $\cos ec\left( x \right)$ is the reciprocal of $\sin \left( x \right)$ and $\cot \left( x \right)$ is the reciprocal of $\tan \left( x \right)$ . In addition to $\cos ec\left( x \right)$ and \$ \cot \le