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

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Solution

Sec2A=cosec(A-42°)
$\Rightarrow$ cosec (90° - 2A) = cosec (A - 42°)
[∵, cosec(90°- θ) = sec θ]
(1 Mark)

$\Rightarrow$ 90° - 2A = A - 42°

$\Rightarrow$ -2A - A = -42° - 90°

(0.5 mark)

$\Rightarrow$ -3A=-132°

$\Rightarrow$ A= $\frac{132}{3}$

$\Rightarrow$ A = 44°

(0.5 mark)

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