Question

If cosec 4A = sec 5A, find A, where 5A is an acute angle.

Answer 1

As 5A is acute, 4A is also an acute angle so that complementary angle relations can be applied on these two angles.

The given equation is transformed to,

$cosec4A=\sec ({90}^{\circ }-4A)=\sec 5A$

So,

$\sec ({90}^{\circ }-4A)=\sec 5A$

$({90}^{\circ }-4A)=5A$,

Or, $9A={90}^{\circ }$

Or $\angle A={10}^{\circ }$