Question

Simplify $\frac{sin (-\frac{11\pi}{3}) tan (\frac{35\pi}{6}) sec (-\frac{7\pi}{3})}{cos (\frac{5\pi}{4}) cosec (\frac{7\pi}{4}) cos (\frac{17\pi}{6})}$

Answer 1

{sin(-11π/3).tan(35π/6)sec(-7π/3)}/{cos(5π/4).cosec(7π/4).cos(17π/6)}

sin(-11π/3) = - sin(11π/3)
= -sin(4π - π/3) = sin(π/3) = √3/2

tan(35π/6) = tan(3 × 2π - π/6) = -tan(π/6) = -1/√3

sec(-7π/3) = sec(7π/3)
= sec(2π + π/3) = sec(π/3) = 2

cos(5π/4) = cos(π + π/4) = - cos(π/4) = -1/√2

cosec(7π/4) = cosec(2π - π/4)
= -cosec(π/4) = -√2

cos(17π/6) = cos(2π + 5π/6)
= cos(5π/6) = cos(π + π/6) = -cos(π/6)
= -√3/2


now , {sin(-11π/3).tan(35π/6)sec(-7π/3)}/{cos(5π/4).cosec(7π/4).cos(17π/6)}

= {(√3/2).(-1/√3).(2)}/{(-1/√2).(-√2).(-√3/2)}

= -1/{-√3/2}

= 2/√3

Answer 2

HELLO DEAR,

Answer: 2/√3

Step-by-step explanation:

{sin(-11π/3)tan(35π/6)sec(-7π/3)}/{cos(5π/4)cosec(7π/4)cos(17π/6)}

therefore,

[sin(-11π/3) = √3/2 ]

[tan(35π/6) = -1/√3]

[sec(-7π/3) = sec(π/3) = 2]

[cos(5π/4) = cos(π + π/4) = - cos(π/4) = -1/√2]

[cosec(7π/4) = cosec(2π - π/4)

= -cosec(π/4) = -√2]

[cos(17π/6) = cos(2π + 5π/6)

= cos(π + π/6) = -cos(π/6) = -√3/2]

now , {sin(-11π/3)tan(35π/6)sec(-7π/3)}/{cos(5π/4)cosec(7π/4)cos(17π/6)}

=> {(√3/2)(-1/√3)(2)}/{(-1/√2)(-√2)(-√3/2)}

=> -1/{-√3/2}

=> 2/√3

I HOPE IT'S HELP YOU DEAR,

THANKS