Question

tan 11 pi/3 - 2 sin 4 pi/6 - 3/4 cosec × cosec pi/4 + 4 cos × cos 17pi/6 = 3-4 under root 3 / 2 : prove that..​

Answer 1

Answer:

Tan ( 11π/3) - 2Sin(4π/6) -(3/4)Cosec²(π/4) + 4Cos²(17π/6) = ( 3 - 4√3) /2

Step-by-step explanation:

tan 11 pi/3 - 2 sin 4 pi/6 - 3/4 cosec × cosec pi/4 + 4 cos × cos 17pi/6 = 3-4 under root 3 / 2

Solving term individually

Tan ( 11π/3) = Tan((12π - π)/3) = Tan (4π - π/3) = -Tan(π/3) = - Tan60° = -√3

Sin(4π/6) = Sin(2π/3) = Sin(π - π/3) = Sin(π/3) = Sin60° = √3 / 2

Cosec²(π/4) =  1/Sin²(π/4) = 1/Sin²(45°) = 1/(1/√2)²  = 2

Cos²(17π/6) = Cos²(3π - π/6) = Cos²(π - π/6) = (-Cos (π/6))² = (-Cos30°)² = (-√3 /2 )² = 3/4

putting all these values

LHS

-√3 - 2√3 / 2  - (3/4)2  + 4 (3/4)

= -√3 - √3   - 3/2  +  3

= 3/2  - 2√3

= 3/2 - 4√3 /2

= ( 3 - 4√3) /2

= RHS

QED

Tan ( 11π/3) - 2Sin(4π/6) -(3/4)Cosec²(π/4) + 4Cos²(17π/6) = ( 3 - 4√3) /2

Answer 2

it is so difficult plz try to understand properly