Trigonometric Ratios of Complementary Angles
Important identities:
1. sin(90 – θ) = cosθ
2. cos(90-θ) = sinθ
3. tan(90-θ) = cotθ
4. cot(90-θ) = secθ
5. sec(90-θ) = cosecθ
6. cosec(90-θ) = secθ
Q.1 Evaluate the following:
(i) sin18°/cos72°
Solution:
sin18°/cos72°
= sin( 90°- 72° )/cos72° [ sin(90-θ) = cosθ ]
= cos 72° / cos 72°
= 1
(ii) sin 26°/cos 64°
Solution:
sin 26°/cos 64°
= sin ( 90° – 64°)/cos 64°
= cos 64° / cos 64°
= 1
(iii) cos 48° – sin 42°
Solution:
cos 48° – sin 42°
= cos ( 90° – 42° ) – sin 42° [ cos ( 90° – θ ) = sin θ ]
= sin 42° -sin 42°
= 0
(iv) cosec 31° – sec 59°
Solution:
cosec 31° – sec 59°
= cosec (90° – 59°) – sec 59° [ cosec (90° – θ)= sec θ
= sec 59° – sec 59°
= 0
2. Show that:
(i) tan 48°tan 23°tan 42° tan 67° = 1
Solution:
tan 48°tan 23°tan 42° tan 67° = 1
L.H.S. = tan 48°tan 23°tan 42° tan 67°
⇒ L.H.S. = tan (90° – 42°) tan(90° – 67°) tan 42° tan 67°
⇒ L.H.S. = cot 42° cot 67° tan 42° tan 67°[ tan(90° – θ)= cotθ ]
⇒ L.H.S. = cot 42° tan 42° cot 67° tan 67° [ tanθ cotθ = 1 ]
⇒ L.H.S. = 1
⇒ L.H.S. = R.H.S ( Hence Proved )
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
L.H.S. = cos 38° cos 52° – sin 38° sin 52°
⇒ L.H.S. = cos (90°-52°) cos 52° – sin ( 90° – 52°) sin 52°
⇒ L.H.S. = sin 52° cos 52° – cos 52° sin 52°
⇒ L.H.S. = 0
⇒ L.H.S. = R.H.S. (Hence Proved)
Q.3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Given:
tan 2A = cot (A -18°)
2A < 90°
To Find:
Value of A =?
Solution:
tan 2A = cot (A -18°)
We know that cot (90° – θ) = tan θ
⇒ cot(90° – 2A) = cot (A -18°)
⇒ 90° – 2A = A -18°
⇒ – 2A – A = -18° – 90°
⇒ – 3A = -108°
⇒ A = -108°/-3
⇒ A = 36°
justification
2A < 90°
2×36° < 90°
⇒ 72° < 90°
Hence Value of A = 36°
Q.4 If tan A = cot B, Prove that A + B = 90°.
Given:
tan A = cot B
To Prove:
A + B = 90°
Proof:
tan A = cot B
tan A = tan (90° – B) [ tan (90° – θ) = cot θ]
⇒ A = 90° – B
⇒ A + B = 90° Hence Proved
Q.5 If sec 4A = cosec (A – 20º), where 4A is an acute angle, find the value of A.
Given:
sec 4A = cosec (A – 20°)
4A < 90°
To Find:
Value of A =?
Solution:
sec 4A = cosec (A – 20°)
⇒ cosec(90°- 4A) = cosec (A – 20°) [ cosec(90° – θ) = secθ ]
90° – 4A = A – 20°
⇒ -4A -A = -20° – 90°
⇒ -5A = -110º
⇒ A = -110°/-5
⇒ A = 22°
Justification:
4A < 90°
⇒ 4×22° < 90°
⇒ 88° < 90°
Hence the value of A = 22°
Q.6 If A, B, and C are interior angles of a triangle ABC , then show that:
sin ( B+C /2 ) = cos A/2.
Given:
In triangle ABC
A, B and C are interior angles
To Show:
sin ( B+C /2 ) = cos A/2
Proof:
We know that the sum of all interior angles of triangle is 180°
A + B + C = 180°
⇒ B + C = 180° – A
⇒ (B + C)/2 = (180° – A)/2
⇒ (B + C)/2 = 180º/2 – A/2
⇒ (B + C)/2 = 90° – A/2
⇒ sin(B+C /2) = sin(90° – A/2)
⇒ sin(B+C /2) = cos A/2 [ sin(90° – θ) = cosθ ]
Q.7 Express sin 67° + cos 75º in terms of trigonometric ratios of angles between 0° and 45° .
Solution:
sin 67° + cos 75º
⇒ sin (90° – 23°) + cos (90° – 15°)
{ we know that sin (90° – θ) = cosθ and cos (90° – θ) = sin θ }
⇒ cos 23° + sin 15°