Signs of Trigonometric Functions

Signs of Trigonometric Functions in Different Quadrants:

θ lies in Quadrant → I II III IV
Trigonometric Functions↓
sin θ + ve + ve – ve – ve
cos θ + ve – ve – ve + ve
tan θ + ve – ve + ve – ve
cosec θ + ve + ve – ve – ve
sec θ + ve – ve – ve + ve
cot θ + ve – ve + ve – ve

Examples Based on Signs of Trigonometric Functions:

Example – 01:

  • State the signs of trigonometric functions
  • sin 675o

sin 675 o = sin (360o + 315o) = sin (315o) = = sin (270o + 45o)

Thus the angle 675o lies in the fourth quadrant, where sin function is negative.



Hence sin 675 o is negative

  • sin 159o

sin 159 o = sin (90o + 69o)

Thus the angle 159o lies in second quadrant, where sin function is positive.

Hence sin 159 o is positive



  • cos 573 o

cos 573 o = cos (360o + 213o) = cos (213o) = = sin (180o + 33o)

Thus the angle 573o lies in the third quadrant, where cos function is negative.

Hence cos 573 o is negative

  • cos 250 o

cos 250 o = cos (180o + 70o)

Thus the angle 250o lies in the third quadrant, where cos function is negative.



Hence cos 250 o is negative

  • cos (-8π/3)c

cos(-8π/3)c = cos (-2π – 2π/3) = cos (- 2π/3) =  cos (- π/2 – π/6)

Thus the angle (-8π/3)c lies in the third quadrant, where cos function is negative.

Hence cos(-8π/3)c is negative

  • cos (11π/9)c

cos(11π/9)c = cos (π + 2π/9)



Thus the angle (11π/9)c lies in the third quadrant, where cos function is negative.

Hence cos(11π/9)c is negative

  • sec (3π/5)c

sec(3π/5)c = sec (π/2 + π/10)

Thus the angle (3π/5)c lies in the second quadrant, where sec function is negative.

Hence sec(3π/5)c is negative



Example – 02:

  • Determine the quadrant in which θ lies from following trigonometric functions.
  • sin θ > 0 and sec θ < 0

Given, sin θ > 0 and sec θ < 0. i.e. sin θ is positive and sec θ is negative

Hence θ lies in the second quadrant.

  • cos θ < 0 and cot θ > 0

Given, cos θ < 0 and cot θ > 0, i.e. cos θ is negative and cot θ is positive

Hence θ lies in the third quadrant.



  • sec θ > 0 and cosec θ < 0

Given, sec θ > 0 and cosec θ < 0, i.e. sec θ is positive and cosec θ is negative

Hence θ lies in the fourth quadrant.

Example – 04:

  • If tan θ= -2 and θ lies in second quadrant, find values of other trigonometric functions.
  • Solution:

Given tan θ = -2

We have, 1+ tan2θ = sec2θ

∴  1+ (-2)2 = sec2θ

∴  5 = sec2θ



∴  sec θ = ± 5

Now θ lies in second quadrant where sec ratio is negative.

∴  sec θ = – 5

∴  cos θ = – 1/5



Now, tan θ = sin θ / cos θ

∴  sin θ = tan θ x cos θ = (-2) x ( – 1/5)  = 2/5

Now, cot  θ = 1/ tan θ = 1/-2 = – 1/2

cosec θ = 1/sin θ = 1/(2/5) = 5/2

Example – 05:

  • If tan θ= 5/12 and π < θ < 3π/2, find values of other trigonometric functions.
  • Solution:

Given tan θ = 5/12 and π < θ < 3π/2

We have, 1+ tan2θ = sec2θ



∴  1+ (5/12)2 = sec2θ

∴  1 + 25/144 = sec2θ

∴  (144 + 25)/144 = sec2θ

∴ 169/144 = sec2θ

∴  sec θ = ± 13/12

Now π < θ < 3π/2, hence θ lies in the third quadrant where sec ratio is negative.

∴  sec θ = – 13/12

∴  cos θ =1/(- 13/12) = – 12/13

Now, tan θ = sin θ / cos θ

∴  sin θ = tan θ x cos θ = (5/12) x ( – 12/13)  =  – 5/13

Now, cot  θ = 1/ tan θ = 1/( 5/12 )= 12/5

cosec θ = 1/sin θ = 1/(- 5/13) = – 13/5

Example – 06:

  • If cos θ= 4/5 and 3π/2 < θ < 2π, find values of other trigonometric functions.
  • Solution:

Given cos θ= 4/5 and 3π/2 < θ < 2π

We have, sin2θ + cos2θ = 1

∴  sin2θ + (4/5)2 = 1

∴  sin2θ = 1 – 16/25

∴  sin2θ = (25 – 16)/25

∴  sin2θ = 9/25

∴  sin θ = ± 3/5

Now 3π/2 < θ < 2π, hence θ lies in the fourth quadrant where sin ratio is negative.

∴  sin θ = – 3/5

Now, tan θ = sin θ / cos θ

∴  tan θ = (- 3/5) / (4/5) = – 3/4

Now, cosec θ = 1/sin θ = 1/(- 3/5) = – 5/3

sec θ = 1/cos θ = 1/(4/5) = 5/4

cot  θ = 1/ tan θ = 1/( -3/4 )= – 4/3

Example – 07:

  • If tan θ=  – 4/3 and 3π/2 < θ < 2π, find values of 3 sec θ + 5 tan θ.
  • Solution:

Given tan θ = – 4/3 and 3π/2 < θ < 2π

We have, 1+ tan2θ = sec2θ

∴  1+ (-4/3)2 = sec2θ

∴  1 + 16/9 = sec2θ

∴  (9 + 16)/9 = sec2θ

∴ 25/9 = sec2θ

∴  sec θ = ± 5/3

Now 3π/2 < θ < 2π, hence θ lies in the fourth quadrant where sec ratio is positive.

∴  sec θ = 5/3

Now, 3 sec θ + 5 tan θ = 3 x (5/3) + 5 x (-4/3) = 5 – 20/3 = (15 – 20)/3 = – 5/3

Example – 08:

  • If sec θ= 2 and 3π/2 < θ < 2π, find values of

  • Solution:

Given sec θ= 2 and 3π/2 < θ < 2π

We have,1 + tan2θ = sec2θ

∴  1 + tan2θ = (2)2

∴  1 + tan2θ = 2

∴  tan2θ = 1

∴  tan θ = ± 1

Now 3π/2 < θ < 2π, hence θ lies in the fourth quadrant where tan ratio is negative.

∴  tan θ = – 1

Now cot θ = 1/tan θ = 1 /(-1) = -1

cos θ = 1/sec θ = 1/2

tan θ = sin θ / cos θ

∴  sin θ = tan θ x cos θ = (-1) x ( 1/2)  =  – 1/2

cosec θ = 1/sin θ = 1/(- 1/2) = – √2

Example – 09:

  • If cos θ= – 3/5 and π < θ < 3π/2, find value of

  • Solution:

Given cos θ= – 3/5 and π < θ < 3π/2

We have, sin2θ + cos2θ = 1

∴  sin2θ + (- 3/5)2 = 1

∴  sin2θ = 1 – 9/25

∴  sin2θ = (25 – 9)/25

∴  sin2θ = 16/25

∴  sin θ = ± 4/5

Now π/ < θ < 3π/2, hence θ lies in the thir quadrant where sin ratio is negative.

∴  sin θ = – 4/5

Now, tan θ = sin θ / cos θ

∴  tan θ = (- 4/5) / (- 3/5) = 4/3

Now, cosec θ = 1/sin θ = 1/(- 4/5) = – 5/4

sec θ = 1/cos θ = 1/(- 3/5) = – 5/3

cot  θ = 1/ tan θ = 1/( 4/3 )= 3/4

Example – 10:

  • If  5 tan A = 7  where π < A  < 3π/2 and sec B =  11  where 3π/2 < B  < 2π, find value of cosec A – tan B.
  • Solution:

Given 5 tan A = 7  where π < A  < 3π/2

∴ tan A = 7/5

∴ cot A = 5/7

We have,1 + cot2A = cosec2A

∴  1 + (5/7)2 = cosec2A

∴  1 + (25/7) = cosec2A

∴  (7 + 25)/7 = cosec2A

∴  cosec2A = 32/7

∴ cosec A = ±  32/√ 7 = ± 42/√ 7

Now π < A < 3π/2, hence A lies in the third quadrant where cosec ratio is negative.

  cosec A = – 42/√ 7

Given sec B =  11  where 3π/2 < B  < 2π

We have,1 + tan2B = sec2B

∴  1 + tan2B = (11)2

∴  1 + tan2B = 11

∴  tan2B = 10

∴  tan B = ± 10

Now 3π/2 < B < 2π, hence B lies in the fourth quadrant where tan ratio is negative.

∴  tan B = – 10

The value of the quantity cosec A – tan B =  – 42/√7 – 10

Example – 11:

  • Find value of 4 cos A + 3 cos B if angles A and B lies in second quadrants and

  • Solution:

sin A = 3/5 and sin B = 4/5

We have, sin2A + cos2A = 1

∴ (3/5)2 + cos2A = 1

∴ cos2A = 1 – 9/25

∴  cos2A = (25 – 9)/25

∴  cos2A = 16/25

∴  cosA = ± 4/5

Now A lies in the second quadrant where cos ratio is negative.

∴  cosA = – 4/5

sin B = 4/5

We have, sin2B + cos2B = 1

∴ (4/5)2 + cos2B = 1

∴ cos2B = 1 – 16/25

∴  cos2B = (25 – 16)/25

∴  cos2B = 9/25

∴  cosB = ± 3/5

Now A lies in the second quadrant where cos ratio is negative.

∴  cosA = – 3/5

The value of 4 cos A + 3 cos B = 4 x (- 4/5) + 3 x (- 3/5)= -16/5 – 9/5 = -25/5 = -1

Example – 12:

  • If 2 sin x = 1, π/2 < x < π and 2 cos y = 1, 3π/2 < y < 2π, find the value of

  • Solution:

2sin x = 1, hence sin x = 1/2, π/2 < x < π

We have, sin2x + cos2x = 1

∴ (1/2)2 + cos2x = 1

∴ cos2x = 1 – 1/4

∴  cos2x = (4-1)/4

∴  cos2x = 3/4

∴  cos x = ± 3/2

Now x lies in the second quadrant where cos ratio is negative.

∴  cos x = – 3/2

Now, tan x = sin x/ cos x = (1/2)/(- 3/2) = – 1/√3

2 cos y = 1, hence cos y = 1/2,  3π/2 < y < 2π

We have, sin2y + cos2y = 1

∴ sin2y + (1/2)2 = 1

∴sin2y = 1 – 1/2

∴  sin2y = (2 – 1)/2

∴  sin2y = 1/2

∴  sin y = ±  1/2

Now y lies in the third quadrant where sin ratio is negative.

∴  sin y = –  1/2

Now, tan y = sin y/ cos y = ( –  1/2)/(1/2) = – 1

Trigonometric Functions

Example – 12:

  • If cos A = sin B = -1/3, where π/2 < A < π and  π < B< 3π/2, then find the value of

  • Solution:

Given cos A = -1/3, π/2 < x < π

We have, sin2A + cos2A = 1

∴ sin2A + (-1/3)2 = 1

∴ sin2A = 1 – 1/9

∴  sin2A = (9-1)/9

∴  sin2A = 8/9

∴  sin A = ± 22/3

Now A lies in the second quadrant where sin ratio is positive.

∴  sin A = 22/3

Now, tan A = sin A/ cos A = (22/3) / (- 1/3)= – 2√2

sin B = – 1/3, π < B< 3π/2

We have, sin2B + cos2B = 1

∴(- 1/3)2 + cos2B = 1

∴ cos2B = 1 – 1/9

∴  cos2B = (9 – 1)/9

∴  cos2B = 8/9

∴  cos B = ± 22/3

Now y lies in the second quadrant where cos ratio is negative.

∴  cos B = –  22/3

Now, tan B = sin B/ cos B = ( –  1/3)/(-  22/3) =  1/2√2

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