Question

please answer me!!!!​

Answer 1

Answer

1

Solution

$\frac{ { \cos(20)^{2} + { \cos(70) } }^{2} }{ { \sin(51) }^{2} + { \sin(31) }^{2} }$

$\cos(90 - 20) = \cos(70) = \ \sin(20) \\ sin(90 - 31) = \sin(59) = \cos(31)$

Now on solving Above

$\frac{ { \sin(20) }^{2} + { \cos(20) }^{2} }{ { \sin(31) }^{2} + { \cos(31) }^{2} }$

so

$\frac{1}{1} = 1$

Answer 2

Step-by-step explanation:

see my question Tom bro.....