Question

Find the value of $\frac{({\sin }^2{22}^o+{\sin }^2{68}^o)}{({\cos }^2{22}^o+{\cos }^2{68}^o)}+{\sin }^2{63}^o+\cos {63}^{\circ }\sin {27}^{\circ }$

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

Answer: (C)

$2$

Given

$\frac{({\sin }^2{22}^o+{\sin }^2{68}^o)}{({\cos }^2{22}^o+{\cos }^2{68}^o)}+{\sin }^2{63}^o+\cos {63}^o\sin {27}^o$

$=\frac{{\sin }^2{22}^o+{\sin }^2\left({90}^o-{22}^o\right)}{{\cos }^2{22}^o+{\cos }^2\left({90}^o-{22}^o\right)}+{\sin }^2{63}^o+\cos {63}^o\times \sin \left({90}^o-{63}^o\right)$

$=\frac{{\sin }^2{22}^o+{\cos }^2{22}^o}{{\cos }^2{22}^o+{\sin }^2{22}^o}+{\sin }^2{63}^o+\cos {63}^o\times \cos {63}^o$

( as we know $\sin ({90}^o-\theta )=\cos \theta$ and $\cos ({90}^o-\theta )=\sin \theta )$

$=\frac{1}{1}+{\sin }^2{63}^o+{\cos }^2{63}^o$

$=1+1$

$=2$

$\therefore$ $\frac{({\sin }^2{22}^o+{\sin }^2{68}^o)}{({\cos }^2{22}^o+{\cos }^2{68}^o)}+{\sin }^2{63}^o+\cos {63}^o\sin {27}^o=2$ will be the answer