Question
Without using trigonometric table, evaluate:
$
\left(\frac{\sin 47^{\circ}}{\cos 43^{\circ}}\right)^2-4 \cos ^2 45^{\circ}+\left(\frac{\cos 43^{\circ}}{\sin 47^{\circ}}\right)^2
$

Answer

$
\begin{aligned}
& \left(\frac{\sin 47^{\circ}}{\cos 43^{\circ}}\right)^2-4 \cos ^2 45^{\circ}+\left(\frac{\cos 43^{\circ}}{\sin 47^{\circ}}\right)^2 \\
& \Rightarrow\left(\frac{\sin 47^{\circ}}{\cos 43^{\circ}}\right)^2+\left(\frac{\cos 43^{\circ}}{\sin 47^{\circ}}\right)^2-4\left(\frac{1}{\sqrt{2}}\right)^2 \\
& \Rightarrow\left(\frac{\sin \left(90^{\circ}-43^{\circ}\right)}{\cos 43^{\circ}}\right)^2+\left(\frac{\cos \left(90^{\circ}-47^{\circ}\right)}{\sin 47^{\circ}}\right)^2-4\left(\frac{1}{2}\right) \\
& \Rightarrow\left(\frac{\cos 43^{\circ}}{\cos 43^{\circ}}\right)^2+\left(\frac{\sin 47^{\circ}}{\sin 47^{\circ}}\right)^2-2 \\
& \Rightarrow 1+1-2=0
\end{aligned}
$

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