Question

If $\sec 2A$=$\csc (A$- ${42}^{\circ }$), where $2A$ is an acute angle, find the value of $A$.

Answer 1

Given,

$2A$ is acute and $\sec 2A=\csc (A-{42}^{\circ })$

We have, $\sec \theta =\csc ({90}^{\circ }-\theta )$

$\sec 2A=\csc ({90}^{\circ }-2A)$

$\therefore \csc ({90}^{\circ }-2A)=\csc (A-{42}^{\circ })$

$({90}^{\circ }-2A)=(A-{42}^{\circ })$

${132}^{\circ }=3A$

$3A={132}^{\circ }$

$A=\frac{{132}^{\circ }}{3}$

$\theta ={44}^{\circ }$