# ICSE Solutions for Selina Concise Chapter 25 Complementary Angles Class 9 Maths

### Notes on Complementary Angles

Complementary Angles

When the sum of two angles is 90°, then the angles are known as complementary angles. In other words, if two angles add up to form a right angle, then these angles are referred to as complementary angles. Here we say, that the two angles complement each other

• 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
Main Points in Complementary angles

(a) Two right angles cannot complement each other.

(b) Two obtuse angles cannot complement each other.

(c) Two complementary angles are acute but vice versa is not possible.

Difference between Complementary and Supplementary Angles

Complementary angles: Sum to 90 degrees

Supplementary angles: Sum to 180 degrees

### Exercise 25

1. Evaluate

(i) cos 22°/sin 68°

(ii) tan 47°/cot 43°

(iii) sec 75°/cosec 15°

(iv) (cos 55°/sin 35°) + (cot 35°/tan 55°)

(v) sin2 40° - cos2 50°

(vi) sec218° - cosec272°

(vii) sin 15° cos 15° - cos 75° sin 75°

(viii) sin 42° sin 48° - cos 42° cos 48°

(vii) sin 15° cos 15° - cos 75° sin 75°
= sin(90° - 75°) cos 15° - cos 75° sin(90° - 15°)
= cos 75° cos 15° - cos 75° cos 15°
= 0

(viii) sin 42° sin 48° - cos 42° cos 48°
= sin (90° - 48°) sin48° - cos (90° - 48°) cos 48°
= cos 48° sin 48° -sin 48° cos 48°
= cos 48° sin 48° - cos 48° sin 48°
= 0

2. Evaluate

(i) sin(90°- A) sin A – cos(90° - A) cos A

(ii) sin2 35° - cos2 55°

(iii) (cot 54°/tan 36°) + (tan 20°/cot 70°) – 2

(iv) (2 tan 53°/cot 37°) - (cot 80°/tan 10°)

(v) cos2 25° - sin265° - tan2 45°

(vi) (sin 77°/cos 13°)2 + (cos 77°/sin 13°)2 – 2cos245°

(i) sin(90° - A)sin A - cos (90°  - A) cos A
= cos A sin A - sin A cos A
= 0

(ii)

3. Show that :

(i) tan 10° tan 15° tan 75° tan 80° = 1
(ii) sin 42° sec 48° + cos 42° cosec 48° = 2

(i) L.H.S.
= tan 10° tan 15° tan 75° tan 80°
= tan (90° - 80°) tan(90° - 75°) tan 75° tan 80°
= cot 80° cot 75° tan 75° tan 80°
= (cot 80° tan 80°)(cot 75° tan 75°)
= (1) (1)
= 1
R.H.S

(iii) LHS = sin26/sec 64 + cos 26/cosec 64

= sin26/sec (90-26) + cos 26/cosec (90-26)

= sin26/cosec 26 + cos 26/sec 26

= sin 26×sin 26 + cos 26×cos 26

= sin² 26 + cos²26

= 1

= RHS

4. Express each of following in term of angles between 0 and 45

(i) sin 59° + tan 63°
(ii) cosec 68
°  + cot 72°
(iii) cos 74
° +sec 67°

(i) sin 59° +  tan 63°
= sin(90 - 31)° + tan(90 - 27)°
= cos 31° + cot 27°

(ii) cosec 68° + cot 72°
= cosec(90 - 22)° + cot(90 - 18)°
= sec 22° + tan 18°

(iii) cos 74° + sec 67°
= cos (90-16)° + sec(90 - 23)°
= sin 16° + cosec 23°

5. For triangle ABC Show that

(i) sin[(A+B)/2]  = cos(C/2)
(ii) tan
[(B+C)/2] = cot(A/2)

6. Evaluate :

(i) 3(sin 72°/cos 18°) – (sec 32°/cosec 58°)

(ii) 3 cos 80° cosec 10° + 2sin59° sec 31°

(iii) (sin 80°/cos 10°) + sin 59° sec 31°

(iv) tan (55° - A) – cot (35° + A)

(v) cosec (65° + A) – sec (25° - A)

(vi) 2(tan57°/cot 33°) – (cot 70°/tan 20°) - √2 cos45°

(vii) (cot241°/tan249°) – 2(sin275°/cos215°)

(viii) (cos 70°/sin 20°) + (cos 59°/sin 31°) – 8sin230°

(ix) 14 sin 30° + 6cos 60° - 5 tan 45°

7. A triangle ABC is right angled at B. Find the value of  (sec A sin C - tan A tan C)/sin B .

Since △ABC is a right - angled triangle, right - angled at B,
A + C = 90°

8. In each case given below find the value of angle A where 0° ≤ A ≤ 90°.

(i) sin(90° - 3A).cosec 42° = 1

(ii) cos(90° - 3A).sec77° = 1

or

(ii) cos (90-A).sec 77 =1

⇒ sin A. sec 77 =1

⇒ sin A.1/cos 77 =1

⇒ sin A.1/cos(90-13) =1

⇒ sin A/sin 13 =1

⇒ sin A = sin 13

On comparing,

A = 13