Question

Evaluate:
(i) $\frac{{\sin }^2{63}^{\circ }+{\sin }^2{27}^{\circ }}{{\cos }^2{17}^{\circ }+{\cos }^2{73}^{\circ }}$

(ii) $\sin {25}^{\circ }\cos {65}^{\circ }+\cos {25}^{\circ }\sin {65}^{\circ }$

Answer 1

(i)

We know that $\cos (90-\theta )=\sin \theta ,\sin (90-\theta )=\cos \theta$

$\sin {63}^o=\sin (90-27{)}^o=\cos {27}^o$

And,

$\cos {17}^o=\cos (90-73{)}^o=\sin {73}^o$

$\therefore \frac{{\sin }^2{63}^o+{\sin }^2{27}^o}{{\cos }^2{17}^o+{\cos }^2{73}^o}$ = $\frac{{\cos }^2{27}^o+{\sin }^2{27}^o}{{\cos }^2{73}^o+{\cos }^2{73}^o}$

$=\frac{1}{1}=1$ ...$[\because {\sin }^2\theta +{\cos }^2\theta =1]$

(ii)

We know that $\cos (90-\theta )=\sin \theta ,\sin (90-\theta )=\cos \theta$

$\sin {25}^o=\sin (90-65{)}^o=\cos {65}^o$

$\cos {25}^o=\cos (90-25{)}^o=\sin {65}^o$

$\therefore \sin {25}^o\cos {65}^o+\cos {25}^o\sin {65}^o$

$=\cos {65}^o\cos {65}^o+\sin {65}^o\sin {65}^o$

$={\cos }^2{65}^o+{\sin }^2{65}^o$

$=1$ ...$[\because {\sin }^2\theta +{\cos }^2\theta =1]$