Question

How do you simplify the expression \[\sec x\cot x-\cot x\cos x\]?

Answer 1

Hint: In this problem, we have to simplify the trigonometric expression to its reduced form. For this, we have to know the trigonometric formulas to simplify these types of problems. We can convert every term using the formula and cancel the similar terms to simplify the expression \[\sec x\cot x-\cot x\cos x\].

Complete step by step answer:
We know that the given trigonometric expression is,
\[\sec x\cot x-\cot x\cos x\] ……. (1)
We can now use formula to simplify the expression (1).
We know that the trigonometric formula to rewrite \[\sec x\] and \[\cot x\]into sines and cosines.
\[\sec x=\dfrac{1}{\cos x}\] and \[\cot x=\dfrac{\cos x}{\sin x}\].
Substituting the above formula in expression (1), we get
\[\begin{align}
  & \Rightarrow \sec x\cot x-\cot x\cos x \\
 & \Rightarrow \dfrac{1}{\cos x}\times \dfrac{\cos x}{\sin x}-\dfrac{\cos }{\sin x}\times \cos x \\
\end{align}\]
Now, we can cancel similar terms to simplify it.
\[\Rightarrow \dfrac{1}{\sin x}-\dfrac{{{\cos }^{2}}x}{\sin x}\]
Here, we have same denominator, so we can combine the terms, we get
\[\Rightarrow \dfrac{1-{{\cos }^{2}}x}{\sin x}\]
We also know that \[1-{{\cos }^{2}}x={{\sin }^{2}}x\] substituting this formula in the above step we get,
\[\begin{align}
  & \Rightarrow \dfrac{{{\sin }^{2}}x}{\sin x} \\
 & \Rightarrow \dfrac{\sin x\times \sin x}{\sin x} \\
\end{align}\]
Here, we can cancel similar terms, we get
\[\Rightarrow \sin x\]
Therefore, the simplified solution of \[\sec x\cot x-\cot x\cos x\] is \[\sin x\].

Note:
Students make mistakes in the trigonometric formula part while rewriting sec and cot into sines and cosines. We should know basic trigonometric rules and formulae to solve or simplify these types of problems. Students may also make mistakes while substituting the correct formula for the terms to be substituted. The trigonometric formula part should be understood and concentrated to solve these types of problems. In this problem, we have used several formulae, remember that \[\cot x=\dfrac{\cos x}{\sin x}\], \[\sec x=\dfrac{1}{\cos x}\] and \[1-{{\cos }^{2}}x={{\sin }^{2}}x\]these are some basic formulae used in these types of problems.