Find the other value of A, if $\left( \sec 2A-1 \right)\left( \csc 3A-1 \right)=0$, when one value is ${{0}^{\circ }}$
Last updated date: 21st Mar 2023
•
Total views: 306k
•
Views today: 5.84k
Answer
306k+ views
Hint: Consider $\sec 2A$ and $\csc 3A$ as variables. This will transform the given expression into an equation in 2 variables. Since the product of $\left( \sec 2A-1 \right)$ and $\left( \csc 3A-1 \right)$ is 0, one of them should at least be 0. Use this fact to solve for A.
Complete step by step solution:
We have been given that the product of $\left( \sec 2A-1 \right)$ and $\left( \csc 3A-1 \right)$ is 0. We now assume that $x=\sec 2A$ and $y=\csc 3A$. Thus the expression can be reduced to an equation in x and y given as $\left( x-1 \right)\left( y-1 \right)=0$.
We know that the solution of any equation of the form $\left( x-a \right)\left( y-b \right)=0$ is given by
$x=a$ and $y=b$. Using the same rule in the above equation,
we get $x=1$ and $y=1$.
Thus, $\sec 2A=1$.
Now, we know that the value of $\sec \theta =1$ for $\theta ={{0}^{\circ }}$. Thus, using this in the above result, we get
$\begin{align}
& 2A={{0}^{\circ }} \\
& \Rightarrow A={{0}^{\circ }} \\
\end{align}$
This is the value of A that is already given to us. Thus, the required other value of A comes from the other result that we have, which is $y=1$.
Thus, $\csc 3A=1$
Now, we know that the value of $\csc \theta =1$ for $\theta =\dfrac{\pi }{2}$ or ${{90}^{\circ }}$. Thus, using this in the above result, we get
$\begin{align}
& 3A={{90}^{\circ }} \\
& \Rightarrow A={{30}^{\circ }} \\
\end{align}$
Thus, the required other value of A is given by $A={{30}^{\circ }}$ or $A=\dfrac{\pi }{6}$.
Note: The equations formed, $\sec 2A=1$ and $\csc 3A=1$, can have numerous solutions known as general values of A. But the fact, given to us that one value of A is ${{0}^{\circ }}$ indicates that we are essentially talking about the principal values of A, which are the values lying in between 0 and $\pi $. The question is to be solved keeping this in mind and hence, choosing only the principal value of A.
Complete step by step solution:
We have been given that the product of $\left( \sec 2A-1 \right)$ and $\left( \csc 3A-1 \right)$ is 0. We now assume that $x=\sec 2A$ and $y=\csc 3A$. Thus the expression can be reduced to an equation in x and y given as $\left( x-1 \right)\left( y-1 \right)=0$.
We know that the solution of any equation of the form $\left( x-a \right)\left( y-b \right)=0$ is given by
$x=a$ and $y=b$. Using the same rule in the above equation,
we get $x=1$ and $y=1$.
Thus, $\sec 2A=1$.
Now, we know that the value of $\sec \theta =1$ for $\theta ={{0}^{\circ }}$. Thus, using this in the above result, we get
$\begin{align}
& 2A={{0}^{\circ }} \\
& \Rightarrow A={{0}^{\circ }} \\
\end{align}$
This is the value of A that is already given to us. Thus, the required other value of A comes from the other result that we have, which is $y=1$.
Thus, $\csc 3A=1$
Now, we know that the value of $\csc \theta =1$ for $\theta =\dfrac{\pi }{2}$ or ${{90}^{\circ }}$. Thus, using this in the above result, we get
$\begin{align}
& 3A={{90}^{\circ }} \\
& \Rightarrow A={{30}^{\circ }} \\
\end{align}$
Thus, the required other value of A is given by $A={{30}^{\circ }}$ or $A=\dfrac{\pi }{6}$.
Note: The equations formed, $\sec 2A=1$ and $\csc 3A=1$, can have numerous solutions known as general values of A. But the fact, given to us that one value of A is ${{0}^{\circ }}$ indicates that we are essentially talking about the principal values of A, which are the values lying in between 0 and $\pi $. The question is to be solved keeping this in mind and hence, choosing only the principal value of A.
Recently Updated Pages
If a spring has a period T and is cut into the n equal class 11 physics CBSE

A planet moves around the sun in nearly circular orbit class 11 physics CBSE

In any triangle AB2 BC4 CA3 and D is the midpoint of class 11 maths JEE_Main

In a Delta ABC 2asin dfracAB+C2 is equal to IIT Screening class 11 maths JEE_Main

If in aDelta ABCangle A 45circ angle C 60circ then class 11 maths JEE_Main

If in a triangle rmABC side a sqrt 3 + 1rmcm and angle class 11 maths JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

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

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Epipetalous and syngenesious stamens occur in aSolanaceae class 11 biology CBSE
