bantuinnn dongg........

(sin² 120° . cos² 135°) / (cot 60° . sec 60°)

= ( (sin 120°)² . (cos 135)² ) / ( (cos 60/sin 60) . ( 1/cos 60))

= ( (sin 120)² . ( cos 135)² / ( 1/sin 60)
= ( ( Sin 120)² . (Cos 135)² . (Sin 60))

sin 120
= Sin (180-60)
= Sin 60
= (√3)/2

Cos 135
= Cos (180-45)
= - cos 45
= - (√2)/2

Sin 60 = (√3)/2

-->

((√3)/2)² . ( (-√2)/2)² . (√3)/2
= 3/4 . 2/4 . (√3)/2
= (3/16)√3

A

[tex] \frac{sin^{2}120.cos^{2} 135}{cot60.sec60} \\ = \frac{( \frac{ \sqrt{3} }{2})^{2}. (\frac{ \sqrt{2} }{2} ) ^{2} }{ \frac{ \sqrt{3}}{3}.2 } \\ = \frac{\frac{3}{4}. \frac{1}{2} }{ \frac{2}{3} \sqrt{3} } \\ = \frac{3}{8}. \frac{3}{2 \sqrt{3} } \\ = \frac{9}{16 \sqrt{3} } . \frac{ \sqrt{3} }{ \sqrt{3} } \\ = \frac{9 \sqrt{3} }{48} \\ = \frac{3}{16} \sqrt{3} [/tex]