Question

2cos30cos20-2cos40sin10-2cos20sin10=? Please answer me.

Answer 1

Answer:

Correct answer is -0.15575175777

Step-by-step explanation:

please mark it brainliest

Answer 2

Answer:

ur ans is:

LHS => cot70+4cos70 = tan20+4sin20 [using cot(x) = tan(90-x) and cos(x) = sin(90-x)]

= (sin20+4sin20cos20)/cos20

= (sin20+2sin40)/cos20 [using 2sin(x)cos(x) = sin(2x)]

= (sin20+sin40+sin40)/cos20

= (2sin30cos10+sin40)/cos20 [using sin(C)+sin(D) = 2sin((C+D)/2)cos(C-D)/2)]

= (cos10+cos50)/cos20 [using sin30 = 1/2 ]

= (2cos30cos20)/cos20 [using cos(C)+cos(D) = 2cos((C+D)/2)cos((C-D)/2)]

= 2cos30

= √3 = RHS (answer)

Read more on Brainly.in - https://brainly.in/question/14622891#readmore