Two acute angles are said to be complementary if the sum of their measures is equal to 90° .
sin θ = cos (90° - θ) <==> cos θ = sin (90° - θ)
cosec θ = sec (90° - θ) <==> sec θ = cosec (90° - θ)
tan θ = cot (90° - θ) <==> cot θ = tan (90° - θ)
Example 1 :
Find the value of the following.
(cos 47/sin 43)2 + (sin 72/cos 18)2 - 2cos2 45
Solution :
= (cos 47/sin 43)2 + (sin 72/cos 18)2 - 2cos2 45
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Using cos θ = sin (90 - θ) cos 47 = sin (90 - 47) cos 47 = sin 43 |
Using sin θ = cos (90 - θ) sin 72 = cos (90 - 72) sin 72 = cos 18 |
cos 45 = 1/√2
= (sin 43/sin 43)2 + (cos 18/cos 18)2 - 2 (1/√2)2
= 12 + 12 - 1
= 1
Example 2 :
Find the value of the following.
(cos 70/sin 20)2 + (cos 59/sin 31)2 + cos θ/sin(90-θ) - 8 cos2 60
Solution :
L.H.S :
= (cos 70/sin 20)2 + (cos 59/sin 31)2 + cos θ/sin(90-θ) - 8 cos2 60
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Using cos θ = sin (90 - θ) cos 70 = sin (90 - 70) cos 70 = sin 20 |
Using sin θ = cos (90 - θ) cos 59 = sin (90 - 59) cos 59 = sin 31 |
sin(90 - θ) = cos θ
cos2 60 = (1/2)2 = 1/4
= (sin 20/sin 20)2 + (sin 31/sin 31)2 + (cos θ/cosθ) - 8(1/4)
= 1 + 1 + 1 - 2
= 3 - 2
= 1
Example 3 :
Find the value of the following.
tan 15° tan 30° tan 45° tan 60° tan 75°
Solution :
= tan 15° tan 30° tan 45° tan 60° tan 75°
tan 75° = cot (90 - 15) = cot 15
tan 30° = 1/√3, tan 45° = 1, tan 60° = √3
= tan 15° (1/√3) 1 (√3) cot 15°
= tan 15° (1/tan 15°)
= 1
Example 4 :
Find the value of the following.
[cot θ / tan (90 - θ)] + [cos (90 - θ) tan θ sec(90 - θ)/sin (90 - θ) cot (90 - θ) cosec (90 - θ)]
Solution :
= [cot θ / tan (90 - θ)] + [cos (90 - θ) tan θ sec(90 - θ)/sin (90 - θ) cot (90 - θ) cosec (90 - θ)]
= [cot θ/cot θ] + [sin θ tan θ cosec θ/cos θ tan θ sec θ]
= 1 + (sin θ cosec θ/cos θ sec θ)
= 1 + 1
= 2
Without using trigonometric tables, evaluate :
Example 5 :
sin 16°/cos 74°
Solution :
= sin 16°/cos 74° ----(1)
sin 16 = cos (90 - 16)
sin 16 = cos 74
Applying the value of sin 16, we get
= cos 74/cos 74
= 1
Example 6 :
sec 11°/cosec 79°
Solution :
= sec 11°/cosec 79° ----(1)
sec 11 = cosec (90 - 11)
sec 11 = cosec 79
Applying the value of sin 11, we get
= cosec 79/cosec 79
= 1
Example 7 :
cos 35°/sin 55°
Solution :
= cos 35°/sin 55° ----(1)
cos 35 = sin (90 - 35)
cos 35 = sin 55
Applying the value of cos 35, we get
= sin 55/sin 55
= 1
Without using trigonometric tables, prove the following :
Example 8 :
cos 81 - sin 9 = 0
Solution :
L.H.S
cos 81 - sin 9 -----(1)
cos 81 = sin (90 - 81)
= sin 9
applying the value of cos 81 in (1), we get
= sin 9 - sin 9
= 0
R.H.S
Hence proved.
Example 9 :
cosec 80 - sec 10 = 0
Solution :
L.H.S
= cosec 80 - sec 10 -----(1)
cosec 80 = sec (90 - 80)
= sec 10
applying the value of cosec 80 in (1), we get
= sec 10 - sec 10
= 0
R.H.S
It is proved.
Example 10 :
cosec2 72 - tan2 18 = 1
Solution :
L.H.S
= cosec2 72 - tan2 18 -----(1)
cosec 72 = sec (90 - 72)
= sec 18
Then, cosec2 72 = sec2 18
Applying the value of cosec2 72 in (1), we get
= sec2 18 - tan2 18
= 1
R.H.S
It is proved.
Example 11 :
cosec2 72 - tan2 18 = 1
Solution :
L.H.S
= cosec2 72 - tan2 18 -----(1)
cosec 72 = sec (90 - 72)
= sec 18
cosec2 72 = sec2 18
Applying the value of cosec2 72 in (1), we get
= sec2 18 - tan2 18
Using sec2 θ - tan2 θ = 1
= 1
Without using trigonometric tables, prove that:
Example 12 :
sin 53° cos 37° + cos 53° sin37° = 1
Solution :
L.H.S
= sin 53° cos 37° + cos 53° sin37°
sin 53 = cos (90 - 53)
sin 53 = cos 37
cos 53 = sin (90 - 53)
cos 53 = sin 37
= cos 37° cos 37° + sin 37° sin 37°
= (cos 37°)2 + (sin 37°)2
= cos237° + sin237°
= 1
R.H.S
Example 13 :
cos 54° cos 36° − sin 54° sin36° = 0
Solution :
L.H.S
= cos 54° cos 36° − sin 54° sin 36°
cos 54° = sin (90 - 54)
= sin 36
sin 54° = cos (90 - 54)
= cos 36
Applying the value of cos 54 and sin 54, we get
= sin 36° cos 36° − cos 36