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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.

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}