Question

$\left\{\frac{{\sin }^{2}22°+{\sin }^{2}68°}{{\cos }^{2}22°+{\cos }^{2}68°}+{\sin }^{2}63°+\cos 63°\sin 27°\right\}$

Answer 1

$$\begin{gathered} \left\{\frac{{\sin }^{2}22°+{\sin }^{2}68°}{{\cos }^{2}22°+{\cos }^{2}68°}+{\sin }^{2}63°+\cos 63°\sin 27°\right\} \\ =\left\{\frac{{\left(\sin \left(90°-68°\right)\right)}^2+{\sin }^{2}68°}{{\cos }^{2}22°+{\cos }^{2}68°}+{\sin }^{2}63°+\cos 63°\sin \left(90°-63°\right)\right\} \\ =\left\{\frac{{\cos }^{2}68°+{\sin }^{2}68°}{{\cos }^{2}22°+{\cos }^{2}68°}+{\sin }^{2}63°+\cos 63°\cos 63°\right\}\left(\because \sin \left(90°-\theta \right)=\cos \theta \right) \\ =\left\{\frac{{\cos }^{2}68°+{\sin }^{2}68°}{{\left(\cos \left(90°-68°\right)\right)}^2+{\cos }^{2}68°}+{\sin }^{2}63°+{\cos }^{2}63°\right\} \\ =\left\{\frac{{\cos }^{2}68°+{\sin }^{2}68°}{{\sin }^{2}68°+{\cos }^{2}68°}+{\sin }^{2}63°+{\cos }^{2}63°\right\}\left(\because \cos \left(90°-\theta \right)=\sin \theta \right) \\ =\left\{1+{\sin }^{2}63°+{\cos }^{2}63°\right\} \\ =\left\{1+1\right\}\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\sin }^2\theta +{\cos }^2\theta =1\right) \\ =2 \\ \mathrm{Hence},\left\{\frac{{\sin }^{2}22°+{\sin }^{2}68°}{{\cos }^{2}22°+{\cos }^{2}68°}+{\sin }^{2}63°+\cos 63°\sin 27°\right\}=2. \end{gathered}$$