Question

Evaluate

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

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

Answer 1

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

$=\frac{{\sin }^2(90-27{)}^{\circ }+{\sin }^2{27}^{\circ }}{{\cos }^2{(90-73)}^{\circ }+{\cos }^2{73}^{\circ }}$ [$\because \sin ({90}^{\circ }-A)=\cos A,\cos ({90}^{\circ }-A)=\sin A$]

$=\frac{{\cos }^2{27}^{\circ }+{\sin }^2{27}^{\circ }}{{\sin }^2{73}^{\circ }+{\cos }^2{73}^{\circ }}$ [$\because {\cos }^2\theta +{\sin }^2\theta =1$]

$=\frac{1}{1}$

$=1$

$\therefore \frac{{\sin }^2{63}^{\circ }+{\sin }^2{27}^{\circ }}{{\cos }^2{17}^{\circ }+{\cos }^2{73}^{\circ }}=1$

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

$=\sin {(90-65)}^{\circ }\cos {65}^{\circ }+\cos {(90-65)}^{\circ }\sin {65}^{\circ }$

$=\cos {65}^{\circ }\cos {65}^{\circ }+\sin {65}^{\circ }\sin {65}^{\circ }$ [$\because \sin ({90}^{\circ }-A)=\cos A,\cos ({90}^{\circ }-A)=\sin A$]

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

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

$\therefore \sin {25}^{\circ }\cos {65}^{\circ }+\cos {25}^{\circ }\sin {65}^{\circ }=1$