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
cos 47 = sin (90 - 47) = sin 43
sin 72 = cos (90 - 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
L.H.S :
= (cos 70/sin 20)2 + (cos 59/sin 31)2 + cos θ/sin(90-θ) - 8 cos2 60
cos 70 = sin (90 - 70) = sin 20
cos 59 = sin (90 - 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
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