Question

2 tan tita=3,then sin square tita+ cos square tita =?

Answer 1

Answer:

thank you for this free point

Answer 2

Answer:

$given \\ \: \: \: 2 \tan \: tita = 3 \\ \tan \: tita = \frac{3}{2} = \frac{opposite \: side}{adjacent \: side } \\ opposite \: side = 3 \: and \: adjacent \: side = 2 \\ {hypotenuse}^{2} = {opposite \: side}^{2} + {adjacent \: side}^{2} \\ {hypotenuse}^{2} = {3}^{2} + {2}^{2} \\ = 9 + 4 = 13 \\ hypotenuse = \sqrt{13} \\ { \sin}^{2} tita + { \cos }^{2} tita = {( \frac{3}{ \sqrt{13} } )}^{2} + {( \frac{2}{ \sqrt{13} } )}^{2} \\ = \frac{9}{13} + \frac{4}{13} = \frac{9 + 4}{13} = \frac{13}{13} \\ = 1 \\ { \sin }^{2} tita \: + { \cos }^{2} \: tita \: = 1$