Question

sin(-660°) tan(1050°) sec(-420°)
$\div$

cos(225°) cosec(315°) cos(570°)

$=$
plz help me​

Answer 1

Answer:

sin(-660°) tan(1050°) sec(-420°)

$\div$

cos(225°) cosec(315°) cos(570°)

$=$

plz help me

Answer 2

S=sin(−660)tan(1050)sec(−420)(cos225)(cosec315)(cos510)

We make the following changes:

sec x=1/cosx

Sec(-420) = [(1/cos(-420)]

cosec x =1 / sin x

Cosec(315) = [(1/sin(315)]

Thus, S=sin(−660)tan(1050)sin(315)cos(225)cos(−420)cos(510)

Now,

sin (-x) = - sin x

sin (-660) = - sin 660

Cos (-x) = cos x

Cos (-420) = cos 420

Thus, S=−sin(660)tan(1050)sin(315)cos(225)cos(420)cos(510)

We further use the identities:

Sin (2π + x ) = sin x

Sin (660) = sin ( 360 + 300) = sin ( 2π + 300) = sin (300)

Sin ( π + x) = - sin x

Sin (300) = sin ( 180 + 120) = sin (π +120) = -(-sin 120 )

Sin (120) = sin ( 180 - 60) = sin 60

- sin(660) = sin (60)

Similarly, Sin (315) = - sin (45)

tan x = sin x / cos x

tan (1050) = sin (1050)/ cos (1050) = sin (2*2π + 330)/cos (2*2π + 330) = sin (330)/cos(330)

Sin ( 330) = sin ( 180 + 150) = sin (π +150) = (-sin 150 )

Sin (150) = sin ( 180 - 30) = - sin 30

Cos( 330) = cos( 180 + 150) = cos(π +150) = (-cos150 )

Cos (150) = cos( 180 - 30) =-(-cos 30) = cos 30

Tan (1050) = tan (30)

Cos(2π + x ) = cos x

Cos(420) =cos (360 +60) = cos (2π +60) =cos 60

Cos (420) =cos (60)

Cos (510) =cos (360 +150) = cos (2π +150) = cos 150

Cos(π-x) = -cos x

Cos ( 150) =cos (180 -30) = cos (π - 30) = - cos (30)

Cos (510) = - cos (30)

Cos ( π + x) = - cos x

Cos (225) = cos ( 180 + 45) = cos (π +45) = - cos (45)

Cos (225) = - cos 45

Thus, S=−sin(660)tan(1050)sin(315)cos(225)cos(420)cos(510)

All the values fall in quadrant 1. So all are positive.

S=sin(60)tan(30)sin(45)cos(45)cos(60)cos(30)

S=(√3/2)∗(1/√3)∗(1/√2)(1/√2)∗(1/2)∗(√3/2)

S=2/√3

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