Question

Evaluate $\frac{{\sin }^2{63}^o+{\sin }^2{27}^o}{{\cos }^2{17}^o+{\cos }^2{73}^o}$.

Answer 1

We know that ${\sin }^2\theta +{\cos }^2\theta =1$

& ${\cos }^{2}A-{\sin }^{2}B=\cos (A+B)\cos (A-B)$

Now, ${\sin }^2{63}^o=1-{\cos }^2{63}^o$

& ${\cos }^2{73}^o=1-{\sin }^2{73}^o$

So,

$\frac{1-{\cos }^2{63}^o+{\sin }^2{27}^o}{{\cos }^2{17}^o+1-{\sin }^2{73}^o}$

$=\frac{1-({\cos }^2{63}^o-{\sin }^2{27}^o)}{1+({\cos }^2{17}^o-{\sin }^2{73}^o)}$

$=\frac{1-\cos (63+27)\cos (63-27{)}^o}{1+\cos (17+73{)}^o\cos (17-73{)}^o}$

$=\frac{1-\cos \left(90{)}^o\cos \right(36{)}^o}{1+\cos \left(90{)}^o\cos \right(-56{)}^o}$

But $\cos \left({90}^o\right)=0$

So,

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