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If the value of \[\text{sec4A=cosec(A-2}{{\text{0}}^{\circ }})\] where 4A is an acute angle then, find the value of A.

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Hint: To solve this question, we will use a basic trigonometric identity, which is given as below; \[\text{sec}\theta \text{=cosec}\left( {{90}^{\circ }}-\theta \right)\] Where $\theta $ is angle. This relation comprises relation between $\text{sec}\theta \text{ and cosec}\theta $ We will apply this to the LHS of the given equation and then compare both sides to get the value of A.

Complete step-by-step solution:
Given that \[\text{sec4A=cosec(A-2}{{\text{0}}^{\circ }})\]
Here, 4A is an acute angle.
We will use a basic trigonometric identity to solve this question. The trigonometric identity is
\[\text{sec}\theta \text{=cosec}\left( {{90}^{\circ }}-\theta \right)\]
As we are given, \[\text{sec4A=cosec(A-2}{{\text{0}}^{\circ }})\]
Here, using above identity in the left hand side, we get
\[\begin{align}
  & \text{sec4A=cosec(9}{{\text{0}}^{\circ }}-4\text{A}) \\
 & \Rightarrow \text{sec4A=cosec(9}{{\text{0}}^{\circ }}-4\text{A})=\text{cosec(A-2}{{\text{0}}^{\circ }}) \\
\end{align}\]
Comparing both the sides, we get:
\[\text{cosec(9}{{\text{0}}^{\circ }}-4\text{A})=\text{cosec(A-2}{{\text{0}}^{\circ }})\]
Now, as both sides has cosec, so we apply $\text{cose}{{\text{c}}^{\text{-1}}}$ on both sides, we get:
\[\begin{align}
  & \text{cose}{{\text{c}}^{\text{-1}}}\left( \text{cosec(9}{{\text{0}}^{\circ }}-4\text{A}) \right)=\text{cose}{{\text{c}}^{\text{-1}}}\left( \text{cosec(A-2}{{\text{0}}^{\circ }}) \right) \\
 & \Rightarrow \text{9}{{\text{0}}^{\circ }}-4\text{A}=\text{A-2}{{\text{0}}^{\circ }} \\
 & \Rightarrow -4\text{A-A=-9}{{\text{0}}^{\circ }}-{{20}^{\circ }} \\
 & \Rightarrow +5\text{A=+11}{{\text{0}}^{\circ }} \\
\end{align}\]
Dividing by 5, we get:
\[\begin{align}
  & \text{A=}\dfrac{{{110}^{\circ }}}{5} \\
 & \text{A=22}^{\circ } \\
\end{align}\]
So, we have value of \[\text{A=22}^{\circ } \]
Therefore, value of \[\text{A=22}^{\circ } \]

Note: Another way to solve this question is using \[\text{sec}\theta \text{=}\dfrac{1}{\text{cos}\theta }\text{ and sin}\theta \dfrac{1}{\text{cosec}\theta }\] trigonometric identity.
We are given \[\text{sec4A=cosec(A-2}{{\text{0}}^{\circ }})\] Applying identity we get \[\dfrac{1}{\text{cos4A}}=\dfrac{1}{\text{sin}\left( A-{{20}^{\circ }} \right)}\Rightarrow \text{cos4A=sin}\left( A-{{20}^{\circ }} \right)\]
Now, we will use trigonometric identity as \[\text{cos}\theta \text{=sin}\left( {{90}^{\circ }}-\theta \right)\Rightarrow \text{sin}\left( {{90}^{\circ }}-4\text{A} \right)=\text{sin}\left( \text{A-2}{{\text{0}}^{\circ }} \right)\]
Applying $\text{si}{{\text{n}}^{-1}}$ both sides, we get:
\[\begin{align}
  & {{90}^{\circ }}-4\text{A=A-2}{{\text{0}}^{\circ }} \\
 & \Rightarrow 5A={{110}^{\circ }} \\
 & \Rightarrow A={{22}^{\circ }} \\
\end{align}\]