Question

Evaluate 3(cos43/sin47)2^-cos37*cosec53/tan5*tan25*tan45*tan65*tan85+sin^2 35/cos^255

Answer 1

Answer:

3

Step-by-step explanation:

3(cos43/sin47)² - cos37*cosec53/tan5*tan25*tan45* tan65*tan 85  + sin² 35/cos²55

//remember sin A = cos (90-A)      cosec A = sec(90-A)

Solving each part separately

(cos43/sin47) => Cos43)/Cos(90-47) = Cos43/Cos43 = 1

cos37*cosec53 = Cos37*Sec(90-53) = Cos37*Sec37 = Cos37*1/Cos37 = 1

//Remember  tan A = cot (90-A) = 1/tan(90-A). Now in denominator part

tan5*tan25*tan45*tan65*tan85 = tan5*tan25*tan45*Cot(90-65)*Cot(90-85) = tan5*tan25*tan45*Cot25*Cot5 = tan5*tan25*tan45*1/tan25*1/tan5

= Tan45 = 1.

Sin²35/Cos²55 = (sin35/cos55)² = [Cos(90-35)/Cos55]² = (Cos55/Cos55)² = 1.

Substituting the values back.

3(1)² - 1/1 + (1)² = 3.