If $\tan 2A=\cot (A-{{18}^{\circ }})$, where $2A$ is an acute angle, then find the value of $A$.
Answer
381k+ views
Hint: Since, $2A$ is an acute angle so we can say that $A$ will also be an acute angle. Thus, all the trigonometric equations applied at this angle will be normal and basic. We will have to convert $\tan 2A$ in the form of the $\cot 2A$ directly, to easily equate both sides of the equation.
“Complete step-by-step answer:”
Here, we have the following given equation as
$\Rightarrow \tan 2A=\cot (A-{{18}^{\circ }})...\text{ (1)}$
We have to convert the $\tan 2A$ in the $\cot 2A$ form.
As per question, $2A$ is an acute angle, then we can say that $A$ will also be an acute angle,
And from trigonometric complementary equations, we have
$\Rightarrow \tan \theta =\cot \left( {{90}^{\circ }}-\theta \right)$
Substituting this value of $\tan \theta $ in terms of $\cot \theta $ in equation (1), we get
\[\begin{align}
& \Rightarrow \tan 2A=\cot \left( {{90}^{\circ }}-2A \right) \\
& \Rightarrow \tan 2A=\cot \left( A-{{18}^{\circ }} \right) \\
\end{align}\]
From both of the above equations, we get
\[\Rightarrow \cot \left( {{90}^{\circ }}-2A \right)=\cot \left( A-{{18}^{\circ }} \right)...\text{ (2)}\]
Thus, comparing the angles from both of the sides of the equation (2), we get
\[\Rightarrow {{90}^{\circ }}-2A=A-{{18}^{\circ }}\]
Transposing terms from either side of the equation to keep similar variables at same side, we get
\[\begin{align}
& \Rightarrow {{90}^{\circ }}-2A=A-{{18}^{\circ }} \\
& \Rightarrow -2A-A=-{{90}^{\circ }}-{{18}^{\circ }} \\
& \Rightarrow -3A=-{{108}^{\circ }} \\
\end{align}\]
Now, cancelling out the negation from both sides, we get
\[\begin{align}
& \Rightarrow -3A=-{{108}^{\circ }} \\
& \Rightarrow 3A={{108}^{\circ }} \\
\end{align}\]
With cross-multiplication in the above equation, we get
\[\begin{align}
& \Rightarrow 3A={{108}^{\circ }} \\
& \Rightarrow A=\dfrac{{{108}^{\circ }}}{3} \\
& \Rightarrow A={{36}^{\circ }} \\
\end{align}\]
Hence, The value of acute angle \[A={{36}^{\circ }}\].
Note: A probable mistake which can be made here is that, while conversion of one of the trigonometric functions to another form, there could be a sign conventional mistake irrespective of the given information about the acute angle.
“Complete step-by-step answer:”
Here, we have the following given equation as
$\Rightarrow \tan 2A=\cot (A-{{18}^{\circ }})...\text{ (1)}$
We have to convert the $\tan 2A$ in the $\cot 2A$ form.
As per question, $2A$ is an acute angle, then we can say that $A$ will also be an acute angle,
And from trigonometric complementary equations, we have
$\Rightarrow \tan \theta =\cot \left( {{90}^{\circ }}-\theta \right)$
Substituting this value of $\tan \theta $ in terms of $\cot \theta $ in equation (1), we get
\[\begin{align}
& \Rightarrow \tan 2A=\cot \left( {{90}^{\circ }}-2A \right) \\
& \Rightarrow \tan 2A=\cot \left( A-{{18}^{\circ }} \right) \\
\end{align}\]
From both of the above equations, we get
\[\Rightarrow \cot \left( {{90}^{\circ }}-2A \right)=\cot \left( A-{{18}^{\circ }} \right)...\text{ (2)}\]
Thus, comparing the angles from both of the sides of the equation (2), we get
\[\Rightarrow {{90}^{\circ }}-2A=A-{{18}^{\circ }}\]
Transposing terms from either side of the equation to keep similar variables at same side, we get
\[\begin{align}
& \Rightarrow {{90}^{\circ }}-2A=A-{{18}^{\circ }} \\
& \Rightarrow -2A-A=-{{90}^{\circ }}-{{18}^{\circ }} \\
& \Rightarrow -3A=-{{108}^{\circ }} \\
\end{align}\]
Now, cancelling out the negation from both sides, we get
\[\begin{align}
& \Rightarrow -3A=-{{108}^{\circ }} \\
& \Rightarrow 3A={{108}^{\circ }} \\
\end{align}\]
With cross-multiplication in the above equation, we get
\[\begin{align}
& \Rightarrow 3A={{108}^{\circ }} \\
& \Rightarrow A=\dfrac{{{108}^{\circ }}}{3} \\
& \Rightarrow A={{36}^{\circ }} \\
\end{align}\]
Hence, The value of acute angle \[A={{36}^{\circ }}\].
Note: A probable mistake which can be made here is that, while conversion of one of the trigonometric functions to another form, there could be a sign conventional mistake irrespective of the given information about the acute angle.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Amniocentesis for sex determination is banned in our class 12 biology CBSE

Trending doubts
The lightest gas is A nitrogen B helium C oxygen D class 11 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Which place is known as the tea garden of India class 8 social science CBSE

What is pollution? How many types of pollution? Define it

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

Give 10 examples for herbs , shrubs , climbers , creepers

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE
