TRIGNOMETRIC RATIOS OF COMPLEMENTARY ANGLES:
If the sum of two angles is 90° then, it is called Complementary Angles. In a right-angled triangle, one angle is 90 °, so the sum of the other two angles is also 90° or they are complementary angles.so the trigonometric ratios of the complementary angles will be
sin (90° – A) = cos A, ; cos (90° – A) = sin A, ; tan (90° – A) = cot A,
cot (90° – A) = tan A, ; sec (90° – A) = cosec A, ; cosec (90° – A) = sec A
Example 1 : Find the value of θ in each of the following. θ is an acute angle.
(i) 3 sec 2θ = 2/√3 (ii) 4 cot 3θ − 4 = 0 (iii) 2 sin 2θ = 1
Solution:
(i) 3 sec 2θ = 2 √3
sec 2θ = (2√3)/3 = 2/√3
We know sec 30° = 2/√3
sec 2θ = sec 30° ⇒ 2θ = 30° ⇒ θ = 15°
(ii) 4 cot 3θ −4 = 0
4 cot 3θ = 4 ⇒ cot 3θ = 1
We know cot 45° = 1
∴ cot 3θ = cot 45° ⇒ 3θ = 45° ⇒ θ = 15°
(iii) 2 sin 2θ = 1
sin 2θ =1/2
We know sin 30° =1/2
∴ sin 2θ = sin 30° ⇒ 2θ = 30° ⇒ θ = 15°
Example 2 : In an acute angle triangle ABC, sin (A + B − C) = 1/2 , cot (A – B + C) = 0 and cos (B + C – A) = 1/2 . What are the values of A, B, and C.
Solution: sin (A + B − C) = 1/2
cot (A − B + C) = 0
cos (B + C – A) = 1/2
We know sin 30° = 1/2;
Given sin (A + B − C) = 1/2
∴ A + B − C = 30° ……(1)
We know cot 90°= 0; Given cot (A – B + C) = 0
∴ A − B + C = 90° ……(2)
We know cos 60° = 1/2; Given cos (B + C – A) = 1/2
∴ B + C – A = 60° ……(3)
Add ① and ②:
(A + B C) + (A – B + C) = 30° + 90°
2A = 120° ⇒ A = 60°
Substitute A = 60° in ① and ③
① → 60° + B – C = 30° → B – C = − 30° …… (4)
③ → B + C – 60° = 60° → B + C = 120° ……(5)
Add equations 4 and 5
(B −C) + (B + C ) = −30° + 120°
2B = 90° ⇒ B = 45°
Substitute B = 45° in ④
B – C = 30°
45° C = 30° → C = 75°
The values are A = 60°, B = 45°and C = 75°
Exercise 8.3
1. Evaluate :
(i) (sin 18°)/(cos 72°) (ii) (tan 26°)/(cot 64°)
(iii) cos 48° – sin 42° (iv) cosec 31° – sec 59°
Solution: (i) (sin 18°)/(cos 72°)
To simplify this, convert the sin function into cos function
We know that, sin 18° is written as cos(90° – 18°), which is equal to the cos 72°.
= sin ((90° −18°))/cos 72°
Substitute the value, to simplify this equation
= cos x 72°/cos x 72° =1
(ii) (tan 26°)/(cot 64°)
To simplify this, convert the tan function into cot function
We know that, tan26° is written as cot(90° – 36°), which is equal to the cot 64°.
= tan(90° −36°)/cot(64) to simplify this equation
= (cot 64°)/(cot 64°) = 1
(iii) cos 48° – sin 42°
To simplify this, convert the cos function into sin function
We know that, cos48° is written as sin(90° – 42°), which is equal to the sin 42°.
= cos (90° – 42°) − sin 42° Substitute the value, to simplify this equation.
= sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59°
To simplify this, convert the cosec function into sec function
We know that, cosec31° is written as sec(90° – 59°), which is equal to the sec 59°
= cosec (90° – 59°) – sec 59° Substitute the value, to simplify this equation
= sec 59° – sec 59° = 0
2. Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Solution:
(i) tan 48° tan 23° tan 42° tan 67°
Simplify the given problem by converting some of the tan functions to the cot functions
We know that, tan 48° = tan (90° – 42°) = cot 42°
tan 23° = tan (90° – 67°) = cot 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
Substitute the values
= cot 42° cot 67° tan 42° tan 67°
= (cot 42° tan 42°) (cot 67° tan 67°) = 1 × 1 = 1
(ii) cos 38° cos 52° – sin 38° sin 52°
Simplify the given problem by converting some of the cos functions to the sin functions
We know that,
cos 38° = cos (90° – 52°) = sin 52°
cos 52°= cos (90°−38°) = sin 38°
= cos (90° – 52°) cos (90°−38°) – sin 38° sin 52°
Substitute the values
= sin 52° sin 38° – sin 38° sin 52° = 0
3.If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution: tan 2A = cot (A− 18°)
We know that tan 2A = cot (90° – 2A)
Substitute the above equation in the given problem
⇒ cot (90° – 2A) = cot (A −18°)
Now, equate the angles,
⇒ 90° – 2A = A −18° ⇒ 108° = 3A
A = (108°)/3
Therefore, the value of A = 36°
4.If tan A = cot B, prove that A + B = 90°.
Solution: tan A = cot B
We know that cot B = tan (90° – B)
To prove A + B = 90°, substitute the above equation in the given problem
tan A = tan (90° – B)
A = 90° – B
A + B = 90°
Hence Proved.
5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution: sec 4A = cosec (A – 20°)
We know that sec 4A = cosec (90° – 4A)
To find the value of A, substitute the above equation in the given problem cosec (90° – 4A) = cosec (A – 20°) Now, equate the angles
90° – 4A = A− 20°
110° = 5A
A = (110°)/5 = 22°
Therefore, the value of A = 22°
6. If A, B and C are interior angles of a triangle ABC, then show that
sin ( B + c/2) = cos a/2
Solution:
We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°
A + B + C = 180° …. (1)
To find the value of ((B + C))/2, simplify the equation (1)
⇒ B + C = 180° – A
⇒ (B + C))/2, = ((180°−A)/2
⇒ (B + C))/2, = (90°−A/2) Now, multiply both sides by sin functions, we get ⇒ sin (B + C)/2 = sin (90°−A/2)
Since sin (90°−A/2) = cos A/2, the above equation is equal to
sin (B + C)/2 = cos A/2
Hence proved.
7.Express sin 67° + cos 75° in terms of trig. ratios of angles between 0° and 45°.
sin 67° + cos 75° (given) In term of sin as cos function and cos as sin function, it can be written as follows sin 67° = sin (90° – 23°)
cos 75° = cos (90° – 15°)
∴ sin 67° + cos 75° = sin (90° – 23°) + cos (90° – 15°) Now,simplify the above equation
= cos 23° + sin 15°
Therefore, sin 67° + cos 75° can also expressed as cos 23° + sin 15°