# Find the other value of A, if $\left( \sec 2A-1 \right)\left( \csc 3A-1 \right)=0$, when one value is ${{0}^{\circ }}$

Answer

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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.

Last updated date: 30th Sep 2023

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