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Question
Prove that:
\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]
Solution
\[LHS = \left( \frac{\sin49°}{\cos41°} \right)^2 + \left( \frac{\cos41°}{\sin49°} \right)^2 \]
\[ = \left( \frac{\cos\left( 90° - 49° \right)}{\cos41°} \right)^2 + \left( \frac{\cos41°}{\cos\left( 90° - 49° \right)} \right)^2 \]
\[ = \left( \frac{\cos41°}{\cos41°} \right)^2 + \left( \frac{\cos41°}{\cos41°} \right)^2 \]
= 12 + 12
= 1 + 1
= 2
= RHS
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