
How do you simplify $ \dfrac{\cos x}{\sec x+\tan x} $ ?
Answer
563.1k+ views
Hint: We will use the relation between trigonometric functions for simplifying the given expression. We will write the terms in the denominator in terms of the sine function and cosine function. Then we will use the identity $ {{\cos }^{2}}x+{{\sin }^{2}}x=1 $ to rewrite the numerator. After that we will factorize the numerator using the identity $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ and obtain the simplification of the given expression.
Complete step by step answer:
We have to simplify the expression $ f\left( x \right)=\dfrac{\cos x}{\sec x+\tan x} $ . Now, using the relations between the trigonometric functions, we know that $ \tan x=\dfrac{\sin x}{\cos x} $ and $ \sec x=\dfrac{1}{\cos x} $ . Substituting these values in the given expression, we get the following,
$ \begin{align}
& f\left( x \right)=\dfrac{\cos x}{\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}} \\
& \therefore f\left( x \right)=\dfrac{\cos x}{\dfrac{1+\sin x}{\cos x}} \\
\end{align} $
We know that $ \dfrac{\left( \dfrac{a}{b} \right)}{\left( \dfrac{c}{d} \right)}=\dfrac{a}{b}\times \dfrac{d}{c} $ . Using this, we can write the above expression as,
$ \begin{align}
& f\left( x \right)=\cos x\times \dfrac{\cos x}{1+\sin x} \\
& \therefore f\left( x \right)=\dfrac{{{\cos }^{2}}x}{1+\sin x} \\
\end{align} $
We have the following identity, $ {{\cos }^{2}}x+{{\sin }^{2}}x=1 $ . Therefore, we can write $ {{\cos }^{2}}x=1-{{\sin }^{2}}x $ . Substituting this value in the numerator, we get the following,
$ f\left( x \right)=\dfrac{1-{{\sin }^{2}}x}{1+\sin x} $
We have the algebraic identity, $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ . Using this identity, we can factorize the numerator as $ 1-{{\sin }^{2}}x=\left( 1+\sin x \right)\left( 1-\sin x \right) $ . Substituting this value in place of the numerator, we get,
$ \begin{align}
& f\left( x \right)=\dfrac{\left( 1+\sin x \right)\left( 1-\sin x \right)}{1+\sin x} \\
& \therefore f\left( x \right)=1-\sin x \\
\end{align} $
Therefore, we obtain the simplification of the given expression as $ 1-\sin x $ .
Note:
It is important to know the relations between trigonometric functions for solving this type of question. We should also know the trigonometric identities as they are useful in the simplification of trigonometric expressions. It is essential that we write each step explicitly while simplifying the given expression. In this way, we can notice the terms or parts of the expression that can be substituted using identities or relations. Also, we can avoid making mistakes and obtain the correct simplification needed.
Complete step by step answer:
We have to simplify the expression $ f\left( x \right)=\dfrac{\cos x}{\sec x+\tan x} $ . Now, using the relations between the trigonometric functions, we know that $ \tan x=\dfrac{\sin x}{\cos x} $ and $ \sec x=\dfrac{1}{\cos x} $ . Substituting these values in the given expression, we get the following,
$ \begin{align}
& f\left( x \right)=\dfrac{\cos x}{\dfrac{1}{\cos x}+\dfrac{\sin x}{\cos x}} \\
& \therefore f\left( x \right)=\dfrac{\cos x}{\dfrac{1+\sin x}{\cos x}} \\
\end{align} $
We know that $ \dfrac{\left( \dfrac{a}{b} \right)}{\left( \dfrac{c}{d} \right)}=\dfrac{a}{b}\times \dfrac{d}{c} $ . Using this, we can write the above expression as,
$ \begin{align}
& f\left( x \right)=\cos x\times \dfrac{\cos x}{1+\sin x} \\
& \therefore f\left( x \right)=\dfrac{{{\cos }^{2}}x}{1+\sin x} \\
\end{align} $
We have the following identity, $ {{\cos }^{2}}x+{{\sin }^{2}}x=1 $ . Therefore, we can write $ {{\cos }^{2}}x=1-{{\sin }^{2}}x $ . Substituting this value in the numerator, we get the following,
$ f\left( x \right)=\dfrac{1-{{\sin }^{2}}x}{1+\sin x} $
We have the algebraic identity, $ {{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right) $ . Using this identity, we can factorize the numerator as $ 1-{{\sin }^{2}}x=\left( 1+\sin x \right)\left( 1-\sin x \right) $ . Substituting this value in place of the numerator, we get,
$ \begin{align}
& f\left( x \right)=\dfrac{\left( 1+\sin x \right)\left( 1-\sin x \right)}{1+\sin x} \\
& \therefore f\left( x \right)=1-\sin x \\
\end{align} $
Therefore, we obtain the simplification of the given expression as $ 1-\sin x $ .
Note:
It is important to know the relations between trigonometric functions for solving this type of question. We should also know the trigonometric identities as they are useful in the simplification of trigonometric expressions. It is essential that we write each step explicitly while simplifying the given expression. In this way, we can notice the terms or parts of the expression that can be substituted using identities or relations. Also, we can avoid making mistakes and obtain the correct simplification needed.
Recently Updated Pages
Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Trending doubts
What is the median of the first 10 natural numbers class 10 maths CBSE

Which women's tennis player has 24 Grand Slam singles titles?

Who is the Brand Ambassador of Incredible India?

Why is there a time difference of about 5 hours between class 10 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE

State and prove converse of BPT Basic Proportionality class 10 maths CBSE

