Question

$\frac{\left({\sin }^{2}23°+{\sin }^{2}67°\right)}{\left({\cos }^{2}13°+{\cos }^{2}77°\right)}+{\sin }^{2}59°+\cos 59°\sin 21°=?$
(a) 3
(b) 2
(c) 1
(d) 0

Answer 1

$$\begin{gathered} \frac{\left({\sin }^{2}23°+{\sin }^{2}67°\right)}{\left({\cos }^{2}13°+{\cos }^{2}77°\right)}+{\sin }^{2}59°+\cos 59°\sin 31° \\ =\left[\frac{{\sin }^{2}23°+{\left(\sin \left(90°-23°\right)\right)}^2}{{\cos }^{2}13°+{\left(\cos \left(90°-13°\right)\right)}^2}\right]+{\sin }^{2}59°+\cos 59°\sin \left(90°-59°\right) \\ =\left[\frac{{\sin }^{2}23°+{\cos }^{2}23°}{{\cos }^{2}13°+{\sin }^{2}13°}\right]+{\sin }^{2}59°+\cos 59°\cos 59°\left(\because \sin \left(90°-\theta \right)=\cos \theta \mathrm{and}\cos \left(90°-\theta \right)=\sin \theta \right) \\ =\left[\frac{1}{1}\right]+{\sin }^{2}59°+{\cos }^{2}59°\left(\mathrm{using}\mathrm{the}\mathrm{identity}:{\sin }^2\theta +{\cos }^2\theta =1\right) \\ =1+1 \\ =2 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{b}\right). \end{gathered}$$


Disclaimer: There must be $\sin 31°$ instead of $\sin 21°$.