Question

If $sec2A=cosec(A-{42}^{\circ })$, where 2A is an acute angle, find the value of A.

Answer 1

$sec2A=cosec(A-{42}^{\circ })$

or, $cosec({90}^o-2A)=cosec(A-{42}^o)$ [∵, cosec(90^o-θ)=secθ]

or, ${90}^o-2A=A-{42}^o$

or, $-2A-A=-{42}^o-{90}^o$

or, $-3A=-{132}^o$

or, $A=\frac{{132}^o}{3}$

or, $A={44}^o$