Important questions
4 Marks Questions
1. In ΔABC and ΔPQR, If 
A and 
P are acute angles such that sin A = sin P then prove that 
A = 
P.
Sol: Given that sin A = sin P
2. Given cot θ = 
, then evaluate
3. In a right angle triangle ABC, right angle is at B. of tan A = 
, then find the value of
i) sin A cos C + cos A sin C
ii) cos A cos C - sin A sin C
4. A chord of a circle of radius 6 cm, is making an angle 60° a the centre. Find the length of the chord.
Sol: Given radius of the circle OA = OB = 6 cm
∠AOB = 60°
OC is hight from 'O' upon AB and it is an angle
bisector then ∠COB = 30°
But the length of the chord AB = 2BC = 2 × 3 = 6 cm
∴ Length of the chord = 6 cm
5. Prove that (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A
Sol: L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= (sin2 A + 2.sin A.cosec A + cosec2A) + (cos2A + 2.cosA.secA + sec2A)
= (sin2 A + cosec2 A + 2) + (cos2 A + sec2 A + 2)
= sin2 A + cos2 A + cosec2 A + sec2 A + 4
= 1 + cosec2 A + sec2 A + 4
6. If cosec θ + cot θ = k then prove that cos θ = 
Given cosec θ + cot θ = k
(cosec θ + cot θ)2 = k2
cosec2 θ + 2 cosec θ. cot θ + cot2 θ = k2
2 Marks Questions
1. The sides of a right angle triangle are PQ = 7 cm, QR = 25 cm and ∠P = 90° respectively. Then find tan Q - tan R.
Sol: Given that PQ = 7 cm, QR = 25 cm and ∠P = 90°
2. If cos A = 
then find sin A and tan A.
3. Show that i) tan 48°. tan 16°. tan 42°. tan 74° = 1
ii) cos 36°. cos 54° - sin 36°. sin 54° = 0
Sol: i) tan 48°. tan 16°. tan 42°. tan 74°
= tan 48°. tan 42°. tan 16°. tan 74°
= tan 48°. tan (90° - 48°). tan 16°. tan (90° - 16°)
= tan 48°. cot 48°. tan 16°. cot 16°
= 1
ii) cos 36°. cos 54° - sin 36°. sin 54°
= cos 36°. cos(90° - 36°) - sin 36°. sin(90° - 36°)
= cos 36°. sin 36° - sin 36°. cos 36°
= 0
4. If A, B and C are interior angles of a traingle ABC, then show that
Sol: Given A, B and C are interior angles of a traiangle ABC,
∴ ∠A + ∠B + ∠C = 180°
∠A + ∠B = 180° - ∠C
5. Evaluate i) (1 + tan θ + sec θ)(1 + cot θ - cosec θ)
ii) (sec2 θ - 1) (cose2 θ - 1)
Sol: i) (1 + tan θ + secθ)(1 + cot θ - cosec θ)
= 2
ii) (sec2 θ - 1)(cose2 θ - 1)
= tan2 θ. cot2 θ (∵ sec2 θ - 1 = tan2 θ and cosec2 θ - 1 = cot2 θ)
= 1
Sol: L.H.S. = (cosec θ - cot θ)2 = cosec2 θ + cot2 θ - 2 cosec θ . cot θ
9. In ΔOPQ, right angle at P, OP = 7 cm, OQ - PQ = 1 cm. Determine the values of sin Q and cos Q.
Sol: In ΔOPQ, we have
OQ2 = OP2 + PQ2
(1 + PQ)2 = OP2 + PQ2 (∵ OQ - PQ = 1 ⇒ OQ = 1 + PQ)
1 + PQ2 + 2 PQ = OP2 + PQ2
10. Prove that (1 + tan A. tan B)2 + (tan A - tan B)2 = sec2 A. sec2 B.
Sol: L.H.S. = (1 + tan A . tan B)2 + (tan A - tan B)2
= 1 + tan2 A . tan2 B + 2 tan A. tan B + tan2 A + tan2 B - 2 tan A tan B
= 1 + tan2 B + tan2 A . tan2 B + tan2 A
= sec2 B + tan2 A (tan2 B + 1)
= sec2 B + tan2 A (sec2 B)
= sec2B (1 + tan2 A)
= sec22 B. sec2 A = R.H.S.
11. In ΔABC, right angle at C and ∠A = ∠B, (i) Is cos A = cos B (ii) Is tan A = tan B?
Sol: We have ∠A = ∠B
⇒ BC = AC
Let BC = AC = x
AB2 = AC2 + BC2
= x2 + x2
= 2x2
1 Mark Questions
2. Is it right to say cos (60° + 30°) = cos 60°.cos 30° - sin 60°.sin 30°
Sol: L.H.S. = cos(60° + 30°)
= cos 90°
= 0
∴ cos(60° + 30°) = cos 60° . cos 30° - sin 60° . sin 30° is right
3. In right angle triangle ΔPQR, right angle is at Q and PQ = 6 cm, ∠RPQ = 60°. Determne the lenghts of QR and PR.
Sol: In ΔPQR, ∠Q = 90°, ∠RPQ = 60°, PQ = 6 cm,
4. In ΔXYZ, right angle is at Y, YZ = x and XZ = 2x then determine ∠YXZ and ∠YZX.
Sol: Given that In ΔXYZ, ∠Y = 90°, YZ = x; XZ = 2x
∴ Z = 60° (i.e) ∠YZX = 60°
∠X = 30° (i.e) ∠YXZ = 30°
= 1
ii) cosec 31° - sec 59° = cosec 31° - sec(90° - 31°)
= cosec 31° - cosec 31°
= 0
6. If tan A = cot B where A and B are acute angles, prove that A + B = 90°.
Sol: Given tan A = cot B
cot (90°- A) = cot B (∵ cot (90° - A) = tanA))
⇒ 90° - A = B
⇒ 90° = A + B
7. Express sin 75° + cos 65° in terms of trigonometric ratios of angles between 0° and 45°.
Sol: Given sin 75° + cos 65° = sin(90° - 15°) + cos(90° - 25°)
= cos15° + sin 25°
8. Evaluate (sin θ + cos θ)2 + (sin θ - cos θ)2
Sol: (sin θ + cos θ)2 + (sin θ - cos θ)2
= (sin2 θ + cos2 θ + 2 sin θ.cos θ) + (sin2 θ + cos2 θ - 2 sin θ.cos θ)
= (1 + 2 sin θ.cos θ) + (1 - 2 sin θ.cos θ)
= 1 + 1 + 2 sin θ.cos θ - sin θ.cos θ
= 2
Writer: T.S.V.S. Suryanarayana Murthy
