Question

Prove that $\frac{({\sin }^{2}73°+{\sin }^{2}17°)}{({\cos }^{2}28°+{\cos }^{2}62°)}=1.$

Answer 1

$\frac{({\sin }^{2}73°+{\sin }^{2}17°)}{({\cos }^{2}28°+{\cos }^{2}62°)}=1.$

$$\begin{gathered} \text{LHS=}\frac{{\sin }^2{73}^0+{\sin }^2{17}^0}{{\cos }^2{28}^0+{\cos }^2{62}^0} \\ \text{=}\frac{{\left[\sin ({90}^0-{17}^0)\right]}^2+{\sin }^2{17}^0}{{\left[\cos ({90}^0-{62}^0)\right]}^2+{\cos }^2{62}^0} \\ \text{=}\frac{{\cos }^2{17}^0+{\sin }^2{17}^0}{{\sin }^2{62}^0+{\cos }^2{62}^0} \\ \text{=}\frac{1}{1}[\because {\sin }^2\theta +{\cos }^2\theta =1] \\ \text{=1=RHS} \end{gathered}$$