Question

sin square 63+ sin square 27 upon cos square 17+cos square 73

Answer 1

sin square 63+ sin square 27 upon cos square 17+cos square 73
=(sin square 63+ sin square 27) ÷{cos square (90-63)+cos square (90-27)}
=(sin square 63+ sin square 27) ÷(sin square 63+ sin square 27)
=1

Answer 2

$\huge\color{red}{\underline{\underline {Question}}}$

$\: \frac{ {sin}^{2}63 + {sin}^{2} 27 }{ {cos}^{2}17 + {cos}^{2}73 }$

$\huge\color{red}{\underline{\underline {Answer}}}$

$We \: know, \\ ☞ \: {sin}^{2}  θ = {cos}^{2} (90 -  θ) \\ ☞ \: {cos}^{2}  θ = {sin}^{2} (90 -  θ) \\ \\ So, \: \frac{ {sin}^{2}63 + {sin}^{2} 27 }{ {cos}^{2}17 + {cos}^{2}73 } \\ \\ ⟹ \: \frac{ {sin}^{2}(90 - 27) + {sin}^{2} 27}{ {cos}^{2}(90 - 73) + {cos}^{2}73 } \\ \\ As, \: {cos}^{2} A + {sin}^{2} A = 1 \\ \\⟹ \: \frac{ {cos}^{2}27 + {sin}^{2} 27}{ {sin}^{2}73 + {cos}^{2} 73 } \\ \\ ⟹ \: \frac{1}{1} \\ ⟹ \: 1$

So, final answer is 1.