Question

please send me answer

Answer 1

$\frac{ { \sin(15) }^{2} + { \sin(75) }^{2} }{ { \cos(36) }^{2} + { \cos(54) }^{2} } \\ \\ = \frac{ { \cos(90 - 15) }^{2} + { \sin(75) }^{2} }{ { \sin(90 - 36) }^{2} + { \cos(54) }^{2} } \\ \\ = \frac{ { \cos(75) }^{2} + { \sin(75) }^{2} }{ { \sin(54) }^{2} + { \cos(54) }^{2} } \\ \\ = \frac{1}{1} \\ \\ = 1$

Answer 2

Given Equation is sin^(25) + sin^2(75)/cos^2(36) + cos^2(54)

= > sin^2(90 - 75) + sin^2(75)/cos^2(90 - 54) + cos^2 (54)

= > cos^2(75) + sin^2(75)/sin^2(54) + cos^2(54)

We know that sin^2A + cos^2A = 1.

= > 1/1

= > 1.

Hope this helps!