Question

Find the value of (3tan41/cot 49)2-(sin35sec55/tan10tan20tan60tan70tan80)²

Answer 1

Answer:

26/3

Step-by-step explanation:

Find the value of (3tan41/cot 49)2-(sin35sec55/tan10tan20tan60tan70tan80)²

$(\frac{3tan41}{Cot49} )^2 - (\frac{Sin35Sec55}{tan10tan20tan60tan70tan80} )^2$

tanθ = Cot(90-θ)

tan41 = Cot(90-41) = Cot49

tan 10 = cot 80

tan 20 = cot 70

Secθ = 1/Cosθ

Sec55 = 1/Cos55

Cosθ = Sin(90-θ)

Cos55 = Sin(90-55) = Sin35

sin35sec55 = Sin35* (1/Sin35) = 1

putting all these values in equation

$= (\frac{3cot49}{cot49})^2 - (\frac{1}{cot80cot70tan60tan70tan80} )^2\\\\= (3)^2 - (\frac{1}{cot80tan80cot70tan70tan60} )^2$

Tanθ.Cotθ = 1

tan80cot80 = 1 & tan70cot70 = 1

$=9 - (\frac{1}{tan60} )^2$

Tan60 = √3

$= 9 - (\frac{1}{\sqrt{3}} )^2\\\\= 9 - \frac{1}{3} \\\\= \frac{27-1}{3} \\\\= \frac{26}{3}$

Answer 2

Answer:

26/3

Step

Find the value of (3tan41/cot 49)2-(sin35sec55/tan10tan20tan60tan70tan80)²

tanθ = Cot(90-θ)

tan41 = Cot(90-41) = Cot49

tan 10 = cot 80

tan 20 = cot 70

Secθ = 1/Cosθ

Sec55 = 1/Cos55

Cosθ = Sin(90-θ)

Cos55 = Sin(90-55) = Sin35

sin35sec55 = Sin35* (1/Sin35) = 1

putting all these values in equation

Tanθ.Cotθ = 1

tan80cot80 = 1 & tan70cot70 = 1

Tan60 = √3