Question

If sec 4 A = cosec (A − 15°), where 4 A is an acute angle, find the value of A.

Answer 1

$$\begin{gathered} \begin{array}{l}\text{sec}4A=\cos ec(A-{15}^0)\\ \end{array} \\ \begin{array}{l}\text{=> cosec}({90}^0-4A)=\cos ec(A-{15}^0)[\because \sec \theta =\cos ec({90}^0-\theta )]\\ \end{array} \\ \begin{array}{l}{\text{=>90}}^0-4A=A-{15}^0\\ \end{array} \\ \begin{array}{l}{\text{=>105}}^0=5A\\ \\ \end{array} \\ \text{=>}A=\frac{{105}^0}{5}={21}^0 \end{gathered}$$