
Prove the following statement.
$\dfrac{\cot A\cos A}{\cot A+\cos A}=\dfrac{\cot A-\cos A}{\cot A\cos A}$
Answer
582k+ views
Hint: In order to prove the given relation, we should have some knowledge regarding the concepts of trigonometry, like the trigonometric formulas, $\cot A=\dfrac{\cos A}{\sin A}$ and ${{\cos }^{2}}A+{{\sin }^{2}}A=1$. Also, we should know the algebraic property that, $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. By using these properties, we will get the desired result.
Complete step-by-step answer:
In this question, we have been asked to prove $\dfrac{\cot A\cos A}{\cot A+\cos A}=\dfrac{\cot A-\cos A}{\cot A\cos A}$. To prove this relation, we will first consider the left hand side of the relation. So, we can write it as,
$LHS=\dfrac{\cot A\cos A}{\cot A+\cos A}$
Now, we know that, $\cot A=\dfrac{\cos A}{\sin A}$. So, we will get the LHS as,
$LHS=\dfrac{\dfrac{\cos A}{\sin A}\cos A}{\dfrac{\cos A}{\sin A}+\cos A}$
We can further write it as,
$LHS=\dfrac{\dfrac{{{\cos }^{2}}A}{\sin A}}{\dfrac{\cos A}{\sin A}+\cos A}$
Now, we will take the LCM of the terms in the denominator. So, we will get,
$LHS=\dfrac{\dfrac{{{\cos }^{2}}A}{\sin A}}{\dfrac{\cos A+\sin A\cos A}{\sin A}}$
We can further write it as,
$LHS=\dfrac{{{\cos }^{2}}A\sin A}{\sin A\left( \cos A+\sin A\cos A \right)}$
We know that common terms of the numerator and the denominator can be cancelled out, so we get,
$LHS=\dfrac{{{\cos }^{2}}A}{\cos A+\sin A\cos A}$
We can see that cos A is common in both the terms of the denominator, so we get,
$LHS=\dfrac{{{\cos }^{2}}A}{\cos A\left( 1+\sin A \right)}$
After cancelling the common term cos A from the numerator and denominator, we will get,
$LHS=\dfrac{\cos A}{\left( 1+\sin A \right)}$
Now, we will multiply the numerator and the denominator by (1 – sin A) to rationalize (1 + sin A). So, we get,
$LHS=\dfrac{\cos A\left( 1-\sin A \right)}{\left( 1+\sin A \right)\left( 1-\sin A \right)}$
Now, we know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. So, we can write $\left( 1+\sin A \right)\left( 1-\sin A \right)=1-{{\sin }^{2}}A$. So, we will get the LHS as,
$LHS=\dfrac{\cos A-\cos A\sin A}{1-{{\sin }^{2}}A}$
And we know that ${{\cos }^{2}}A+{{\sin }^{2}}A=1$. So, we can write $1-{{\sin }^{2}}A={{\cos }^{2}}A$. Therefore, we get,
$LHS=\dfrac{\cos A-\cos A\sin A}{co{{\operatorname{s}}^{2}}A}$
Now, we will divide the numerator and denominator by sin A. So, we will get,$LHS=\dfrac{\dfrac{\cos A-\cos A\sin A}{\sin A}}{\dfrac{{{\cos }^{2}}A}{\sin A}}$
It can be further written as,
\[\begin{align}
& LHS=\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\cos A\sin A}{\sin A}}{\dfrac{\cos A}{\sin A}\cos A} \\
& LHS=\dfrac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{\cos A}{\sin A}\cos A} \\
\end{align}\]
We know that, $\cot A=\dfrac{\cos A}{\sin A}$. So, we can write $\dfrac{\cos A}{\sin A}=\cot A$. Therefore, we get,
$\begin{align}
& LHS=\dfrac{\cot A-\cos A}{\cot A\cos A} \\
& LHS=RHS \\
\end{align}$
Hence proved.
Note: In this question, the possible mistake one can make is by directly rationalizing the LHS denominator without converting them into cos and sin. This is also correct, but this method will be lengthier and complicated.
Complete step-by-step answer:
In this question, we have been asked to prove $\dfrac{\cot A\cos A}{\cot A+\cos A}=\dfrac{\cot A-\cos A}{\cot A\cos A}$. To prove this relation, we will first consider the left hand side of the relation. So, we can write it as,
$LHS=\dfrac{\cot A\cos A}{\cot A+\cos A}$
Now, we know that, $\cot A=\dfrac{\cos A}{\sin A}$. So, we will get the LHS as,
$LHS=\dfrac{\dfrac{\cos A}{\sin A}\cos A}{\dfrac{\cos A}{\sin A}+\cos A}$
We can further write it as,
$LHS=\dfrac{\dfrac{{{\cos }^{2}}A}{\sin A}}{\dfrac{\cos A}{\sin A}+\cos A}$
Now, we will take the LCM of the terms in the denominator. So, we will get,
$LHS=\dfrac{\dfrac{{{\cos }^{2}}A}{\sin A}}{\dfrac{\cos A+\sin A\cos A}{\sin A}}$
We can further write it as,
$LHS=\dfrac{{{\cos }^{2}}A\sin A}{\sin A\left( \cos A+\sin A\cos A \right)}$
We know that common terms of the numerator and the denominator can be cancelled out, so we get,
$LHS=\dfrac{{{\cos }^{2}}A}{\cos A+\sin A\cos A}$
We can see that cos A is common in both the terms of the denominator, so we get,
$LHS=\dfrac{{{\cos }^{2}}A}{\cos A\left( 1+\sin A \right)}$
After cancelling the common term cos A from the numerator and denominator, we will get,
$LHS=\dfrac{\cos A}{\left( 1+\sin A \right)}$
Now, we will multiply the numerator and the denominator by (1 – sin A) to rationalize (1 + sin A). So, we get,
$LHS=\dfrac{\cos A\left( 1-\sin A \right)}{\left( 1+\sin A \right)\left( 1-\sin A \right)}$
Now, we know that $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$. So, we can write $\left( 1+\sin A \right)\left( 1-\sin A \right)=1-{{\sin }^{2}}A$. So, we will get the LHS as,
$LHS=\dfrac{\cos A-\cos A\sin A}{1-{{\sin }^{2}}A}$
And we know that ${{\cos }^{2}}A+{{\sin }^{2}}A=1$. So, we can write $1-{{\sin }^{2}}A={{\cos }^{2}}A$. Therefore, we get,
$LHS=\dfrac{\cos A-\cos A\sin A}{co{{\operatorname{s}}^{2}}A}$
Now, we will divide the numerator and denominator by sin A. So, we will get,$LHS=\dfrac{\dfrac{\cos A-\cos A\sin A}{\sin A}}{\dfrac{{{\cos }^{2}}A}{\sin A}}$
It can be further written as,
\[\begin{align}
& LHS=\dfrac{\dfrac{\cos A}{\sin A}-\dfrac{\cos A\sin A}{\sin A}}{\dfrac{\cos A}{\sin A}\cos A} \\
& LHS=\dfrac{\dfrac{\cos A}{\sin A}-\cos A}{\dfrac{\cos A}{\sin A}\cos A} \\
\end{align}\]
We know that, $\cot A=\dfrac{\cos A}{\sin A}$. So, we can write $\dfrac{\cos A}{\sin A}=\cot A$. Therefore, we get,
$\begin{align}
& LHS=\dfrac{\cot A-\cos A}{\cot A\cos A} \\
& LHS=RHS \\
\end{align}$
Hence proved.
Note: In this question, the possible mistake one can make is by directly rationalizing the LHS denominator without converting them into cos and sin. This is also correct, but this method will be lengthier and complicated.
Recently Updated Pages
Master Class 11 Chemistry: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

How many 5 digit telephone numbers can be constructed class 11 maths CBSE

Draw a well labelled diagram of reflex arc and explain class 11 biology CBSE

What is the difference between noise and music Can class 11 physics CBSE

Trending doubts
1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

In what year Guru Nanak Dev ji was born A15 April 1469 class 11 social science CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Write the differences between monocot plants and dicot class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 ton equals to A 100 kg B 1000 kg C 10 kg D 10000 class 11 physics CBSE

