Question

$\left\{\frac{({\sin }^{2}22°+{\sin }^{2}68°)}{({\cos }^{2}22°+{\cos }^{2}68°)}+{\sin }^{2}63°+\cos 63°\sin 27\right\}=?$
(a) 0
(b) 1
(c) 2
(d) 3

Answer 1

$$\begin{gathered} (c)\; 2 \\ \text{We have:} \\ [\frac{{\sin }^2{22}^0+{\sin }^2{68}^0}{{\cos }^2{22}^0+{\cos }^{2}68}+{\sin }^2{63}^0+\cos {63}^0\sin {27}^0] \\ =\left[\frac{{\sin }^2{22}^0+{\sin }^2({90}^0-{22}^0)}{{\cos }^2({90}^0-{68}^0)+{\cos }^2{68}^0}+{\sin }^2{63}^0+\cos {63}^0\left\{\sin ({90}^0-{63}^0)\right\}\right] \\ =[\frac{{\sin }^2{22}^0+{\cos }^2{22}^0}{{\sin }^2{68}^0+{\cos }^2{68}^0}+{\sin }^2{63}^0+\cos {63}^0\cos {63}^0][\because \sin ({90}^0-\theta )=\cos \theta \text{and}\cos ({90}^0-\theta )=\sin \theta ] \\ =[\frac{1}{1}+{\sin }^2{63}^0+{\cos }^2{63}^0][\because {\sin }^2\theta +{\cos }^2\theta =1] \\ =1+1=2 \end{gathered}$$