Question
If $A$ and $B$ are complementary angles, prove that: $cosec^2A + cosec^2B = cosec^2A \ cosec^2B$
✓
Answer
Since, $A$ and $B$ are complementary angles, $A + B = 90^\circ$
$cosec^2A + cosec^2B$
$= cosec^2A + (cosec(90^\circ - A))^2$
$= cosec^2A + sec^2A$
$=\frac{1}{\sin ^2 A}+\frac{1}{\cos ^2 A}$
$=\frac{\cos ^2 A+\sin ^2 A}{\sin ^2 A \cos ^2 A}$
$=\frac{1}{\sin ^2 A \cos ^2 A}$
$= cosec^2A \ (sec(90A^\circ - B))^2$
$= cosec^2A \ cosec^2B$
$cosec^2A + cosec^2B$
$= cosec^2A + (cosec(90^\circ - A))^2$
$= cosec^2A + sec^2A$
$=\frac{1}{\sin ^2 A}+\frac{1}{\cos ^2 A}$
$=\frac{\cos ^2 A+\sin ^2 A}{\sin ^2 A \cos ^2 A}$
$=\frac{1}{\sin ^2 A \cos ^2 A}$
$= cosec^2A \ (sec(90A^\circ - B))^2$
$= cosec^2A \ cosec^2B$
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