Question

${\left(\frac{\sin 47°}{\cos 43°}\right)}^2+{\left(\frac{\cos 43°}{\sin 47°}\right)}^2-4{\cos }^{2}45°$

Answer 1

$$\begin{gathered} {\left(\frac{\sin 47°}{\cos 43°}\right)}^2+{\left(\frac{\cos 43°}{\sin 47°}\right)}^2-4{\cos }^{2}45° \\ ={\left(\frac{\sin (90-43)°}{\cos 43°}\right)}^2+{\left(\frac{\cos 43°}{\sin (90-43)°}\right)}^2-4{\left(\frac{1}{\sqrt{2}}\right)}^2(\mathrm{From}\mathrm{the}\mathrm{table}) \\ ={\left(\frac{\cos 43°}{\cos 43°}\right)}^2+{\left(\frac{\cos 43°}{\cos 43°}\right)}^2-4\left({\displaystyle \frac{1}{2}}\right)[\because \sin (90-\mathrm{\theta })=\cos \mathrm{\theta }] \\ =1^2+1^2-2=0 \end{gathered}$$