Question

Prove that:

\[\left( \frac{\sin49^\circ}{\cos41^\circ} \right)^2 + \left( \frac{\cos41^\circ}{\sin49^\circ} \right)^2 = 2\]

Answer 1

\[LHS = \left( \frac{\sin49°}{\cos41°} \right)^2 + \left( \frac{\cos41°}{\sin49°} \right)^2 \]

\[ = \left( \frac{\cos\left( 90° - 49° \right)}{\cos41°} \right)^2 + \left( \frac{\cos41°}{\cos\left( 90° - 49° \right)} \right)^2 \]

\[ = \left( \frac{\cos41°}{\cos41°} \right)^2 + \left( \frac{\cos41°}{\cos41°} \right)^2 \]

= 12 + 12

= 1 + 1

= 2

= RHS