Exercise 8.1
1In ΔABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine: (i) sin A, cos A (ii) sin C, cos C
By Pythagoras theorem:
AC2 = AB2 + BC2 = 242 + 72 = 576 + 49 = 625 ⇒ AC = 25 cm
(i) sin A, cos A
sin A = BCAC = 725 ; cos A = ABAC = 2425
sin A = 725, cos A = 2425
(ii) sin C, cos C
sin C = ABAC = 2425 ; cos C = BCAC = 725
sin C = 2425, cos C = 725
2In Fig. 8.13, find tan P – cot R. (ΔPQR is right-angled at Q, PQ = 12 cm and PR = 13 cm.)
QR2 = PR2 – PQ2 = 169 – 144 = 25 ⇒ QR = 5 cm
tan P = QRPQ = 512 ; cot R = QRPQ = 512
tan P – cot R = 512 – 512 = 0
tan P – cot R = 0
3If sin A = 34, calculate cos A and tan A.
Consider ΔABC right-angled at B. sin A = BCAC = 34. Let BC = 3k and AC = 4k.
AB = √AC2 – BC2 = √16k2 – 9k2 = √7k
cos A = ABAC = √7k4k = √74 ; tan A = BCAB = 3k√7k = 3√7
cos A = √74, tan A = 3√7
4Given 15 cot A = 8, find sin A and sec A.
cot A = 815 = ABBC (in ΔABC right-angled at B). Let AB = 8k and BC = 15k.
AC = √(8k)2 + (15k)2 = √289k2 = 17k
sin A = BCAC = 1517 ; sec A = ACAB = 178
sin A = 1517, sec A = 178
5Given sec θ = 1312, calculate all other trigonometric ratios.
sec θ = hypotenuseadjacent side = 1312. Let hypotenuse = 13k and adjacent side = 12k.
Opposite side = √(13k)2 – (12k)2 = √25k2 = 5k
sin θ = 513 ; cos θ = 1213 ; tan θ = 512
cosec θ = 135 ; cot θ = 125
sin θ = 513, cos θ = 1213, tan θ = 512, cosec θ = 135, cot θ = 125
6If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Consider a triangle ABC right-angled at C, so that ∠A and ∠B are acute angles.
cos A = ACAB ; cos B = BCAB
cos A = cos B ⇒ ACAB = BCAB ⇒ AC = BC
Angles opposite to equal sides of a triangle are equal.
So, ∠A = ∠B
7If cot θ = 78, evaluate: (i) (1 + sin θ)(1 – sin θ)(1 + cos θ)(1 – cos θ) (ii) cot2 θ
cot θ = 78. Let adjacent side = 7k and opposite side = 8k.
Hypotenuse = √49k2 + 64k2 = √113k
sin θ = 8√113 ; cos θ = 7√113
(i)
(1 + sin θ)(1 – sin θ)(1 + cos θ)(1 – cos θ) = 1 – sin2 θ1 – cos2 θ = 1 – 64/1131 – 49/113 = 49/11364/113 = 4964
4964
8If 3 cot A = 4, check whether 1 – tan2 A1 + tan2 A = cos2 A – sin2 A or not.
cot A = 43. Let adjacent side = 4k and opposite side = 3k. Hypotenuse = 5k.
tan A = 34 ; sin A = 35 ; cos A = 45
LHS = 1 – 9/161 + 9/16 = 7/1625/16 = 725
RHS = 1625 – 925 = 725
LHS = RHS.
Yes, 1 – tan2 A1 + tan2 A = cos2 A – sin2 A
9In triangle ABC, right-angled at B, if tan A = 1√3, find the value of: (i) sin A cos C + cos A sin C (ii) cos A cos C – sin A sin C
tan A = BCAB = 1√3. Let BC = k and AB = √3k.
AC = √k2 + 3k2 = 2k
sin A = 12, cos A = √32 ; sin C = ABAC = √32, cos C = BCAC = 12
(i)
sin A cos C + cos A sin C = 12 × 12 + √32 × √32 = 14 + 34 = 1
1
(ii)
cos A cos C – sin A sin C = √32 × 12 – 12 × √32 = 0
0
10In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Let QR = x cm. Then PR = (25 – x) cm.
PR2 = PQ2 + QR2 ⇒ (25 – x)2 = 25 + x2 ⇒ 625 – 50x = 25 ⇒ x = 12
QR = 12 cm, PR = 13 cm
sin P = QRPR = 1213 ; cos P = PQPR = 513 ; tan P = QRPQ = 125
sin P = 1213, cos P = 513, tan P = 125
11State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
False
For example, tan 60° = √3, which is greater than 1.
(ii) sec A = 125 for some value of angle A.
True
sec A = hypotenuseadjacent side. The hypotenuse is the longest side, so sec A ≥ 1. Since 125 > 1, such an angle exists.
(iii) cos A is the abbreviation used for the cosecant of angle A.
False
cos A is the abbreviation for the cosine of angle A. Cosecant is written as cosec A.
(iv) cot A is the product of cot and A.
False
cot A is one symbol for the cotangent of angle A. ‘cot’ alone has no meaning.
(v) sin θ = 43 for some angle θ.
False
sin θ = opposite sidehypotenuse, and the hypotenuse is the longest side. So, sin θ cannot be greater than 1.
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Exercise 8.2
1Evaluate the following:
(i) sin 60° cos 30° + sin 30° cos 60°
√32 × √32 + 12 × 12 = 34 + 14 = 1
1
(ii) 2 tan2 45° + cos2 30° – sin2 60°
2(1)2 + (√32)2 – (√32)2 = 2 + 34 – 34 = 2
2
(iii) cos 45°sec 30° + cosec 30°
1/√22/√3 + 2 = 1/√2(2 + 2√3)/√3 = √32√2(1 + √3)
= √3(√3 – 1)2√2(√3 + 1)(√3 – 1) = 3 – √34√2 = 3√2 – √68
3√2 – √68
(iv) sin 30° + tan 45° – cosec 60°sec 30° + cos 60° + cot 45°
1/2 + 1 – 2/√32/√3 + 1/2 + 1 = (3√3 – 4)/2√3(3√3 + 4)/2√3 = 3√3 – 43√3 + 4
= (3√3 – 4)2(3√3)2 – 42 = 27 – 24√3 + 1627 – 16 = 43 – 24√311
43 – 24√311
(v) 5 cos2 60° + 4 sec2 30° – tan2 45°sin2 30° + cos2 30°
5(1/4) + 4(4/3) – 11/4 + 3/4 = 54 + 163 – 1 = 15 + 64 – 1212 = 6712
6712
2Choose the correct option and justify your choice:
(i) 2 tan 30°1 + tan2 30° = (A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
2/√31 + 1/3 = 2√3 × 34 = √32 = sin 60°
(A) sin 60°
(ii) 1 – tan2 45°1 + tan2 45° = (A) tan 90° (B) 1 (C) sin 45° (D) 0
(D) 0
(iii) sin 2A = 2 sin A is true when A = (A) 0° (B) 30° (C) 45° (D) 60°
For A = 0°: sin 0° = 0 and 2 sin 0° = 0
(A) 0°
(iv) 2 tan 30°1 – tan2 30° = (A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
2/√31 – 1/3 = 2√3 × 32 = √3 = tan 60°
(C) tan 60°
3If tan (A + B) = √3 and tan (A – B) = 1√3; 0° < A + B ≤ 90°; A > B, find A and B.
tan (A + B) = √3 = tan 60° ⇒ A + B = 60° ... (1)
tan (A – B) = 1√3 = tan 30° ⇒ A – B = 30° ... (2)
Adding (1) and (2): 2A = 90°, so A = 45°. Then B = 15°.
A = 45°, B = 15°
4State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
False
Take A = 60° and B = 30°. sin 90° = 1, but sin 60° + sin 30° = √3 + 12 ≠ 1.
(ii) The value of sin θ increases as θ increases.
True
sin 0° = 0, sin 30° = 12, sin 45° = 1√2, sin 60° = √32, sin 90° = 1. (For 0° ≤ θ ≤ 90°)
(iii) The value of cos θ increases as θ increases.
False
cos 0° = 1, cos 30° = √32, cos 60° = 12, cos 90° = 0. It decreases.
(iv) sin θ = cos θ for all values of θ.
False
It is true only for θ = 45°. For example, sin 30° = 12 but cos 30° = √32.
(v) cot A is not defined for A = 0°.
True
cot 0° = cos 0°sin 0° = 10, which is not defined.
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Exercise 8.3
1Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
cosec2 A = 1 + cot2 A ⇒ sin A = 1cosec A = 1√1 + cot2 A
sec A = 1cos A = 1sin A × cot A = √1 + cot2 Acot A
tan A = 1cot A
sin A = 1√1 + cot2 A, sec A = √1 + cot2 Acot A, tan A = 1cot A
2Write all the other trigonometric ratios of ∠A in terms of sec A.
cos A = 1sec A
tan2 A = sec2 A – 1 ⇒ tan A = √sec2 A – 1
sin A = tan A × cos A = √sec2 A – 1sec A
cot A = 1tan A = 1√sec2 A – 1
cosec A = 1sin A = sec A√sec2 A – 1
3Choose the correct option. Justify your choice.
(i) 9 sec2 A – 9 tan2 A = (A) 1 (B) 9 (C) 8 (D) 0
9(sec2 A – tan2 A) = 9 × 1 = 9
(B) 9
(ii) (1 + tan θ + sec θ)(1 + cot θ – cosec θ) = (A) 0 (B) 1 (C) 2 (D) –1
(cos θ + sin θ + 1cos θ)(sin θ + cos θ – 1sin θ) = (sin θ + cos θ)2 – 1sin θ cos θ
= 1 + 2 sin θ cos θ – 1sin θ cos θ = 2
(C) 2
(iii) (sec A + tan A)(1 – sin A) = (A) sec A (B) sin A (C) cosec A (D) cos A
1 + sin Acos A × (1 – sin A) = 1 – sin2 Acos A = cos2 Acos A = cos A
(D) cos A
(iv) 1 + tan2 A1 + cot2 A = (A) sec2 A (B) –1 (C) cot2 A (D) tan2 A
sec2 Acosec2 A = 1/cos2 A1/sin2 A = sin2 Acos2 A = tan2 A
(D) tan2 A
4Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(i) (cosec θ – cot θ)2 = 1 – cos θ1 + cos θ
LHS = (1sin θ – cos θsin θ)2 = (1 – cos θ)2sin2 θ = (1 – cos θ)21 – cos2 θ
= (1 – cos θ)2(1 – cos θ)(1 + cos θ) = 1 – cos θ1 + cos θ = RHS
Hence proved.
(ii) cos A1 + sin A + 1 + sin Acos A = 2 sec A
LHS = cos2 A + (1 + sin A)2(1 + sin A) cos A = cos2 A + 1 + 2 sin A + sin2 A(1 + sin A) cos A
= 2 + 2 sin A(1 + sin A) cos A = 2(1 + sin A)(1 + sin A) cos A = 2cos A = 2 sec A = RHS
Hence proved.
(iii) tan θ1 – cot θ + cot θ1 – tan θ = 1 + sec θ cosec θ
Write s = sin θ and c = cos θ.
LHS = s/c(s – c)/s + c/s(c – s)/c = s2c(s – c) – c2s(s – c) = s3 – c3sc(s – c)
= (s – c)(s2 + sc + c2)sc(s – c) = 1 + scsc = 1 + 1sc = 1 + sec θ cosec θ = RHS
Hence proved.
(iv) 1 + sec Asec A = sin2 A1 – cos A
LHS = 1sec A + 1 = cos A + 1
RHS = 1 – cos2 A1 – cos A = (1 – cos A)(1 + cos A)1 – cos A = 1 + cos A
LHS = RHS.
Hence proved.
(v) cos A – sin A + 1cos A + sin A – 1 = cosec A + cot A, using the identity cosec2 A = 1 + cot2 A.
Divide the numerator and the denominator by sin A.
LHS = cot A – 1 + cosec Acot A + 1 – cosec A = (cot A + cosec A) – (cosec2 A – cot2 A)cot A – cosec A + 1
= (cosec A + cot A) – (cosec A + cot A)(cosec A – cot A)cot A – cosec A + 1
= (cosec A + cot A)(1 – cosec A + cot A)cot A – cosec A + 1 = cosec A + cot A = RHS
Hence proved.
(vi) √1 + sin A1 – sin A = sec A + tan A
LHS = √(1 + sin A)(1 + sin A)(1 – sin A)(1 + sin A) = √(1 + sin A)21 – sin2 A = √(1 + sin A)2cos2 A
= 1 + sin Acos A = 1cos A + sin Acos A = sec A + tan A = RHS
Hence proved.
(vii) sin θ – 2 sin3 θ2 cos3 θ – cos θ = tan θ
LHS = sin θ (1 – 2 sin2 θ)cos θ (2 cos2 θ – 1) = sin θ (1 – 2 sin2 θ)cos θ (2(1 – sin2 θ) – 1)
= sin θ (1 – 2 sin2 θ)cos θ (1 – 2 sin2 θ) = sin θcos θ = tan θ = RHS
Hence proved.
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
LHS = sin2 A + cosec2 A + 2 sin A cosec A + cos2 A + sec2 A + 2 cos A sec A
= (sin2 A + cos2 A) + 2 + 2 + cosec2 A + sec2 A
= 1 + 4 + (1 + cot2 A) + (1 + tan2 A) = 7 + tan2 A + cot2 A = RHS
Hence proved.
(ix) (cosec A – sin A)(sec A – cos A) = 1tan A + cot A
LHS = (1 – sin2 Asin A)(1 – cos2 Acos A) = cos2 Asin A × sin2 Acos A = sin A cos A
RHS = 1sin A/cos A + cos A/sin A = sin A cos Asin2 A + cos2 A = sin A cos A
LHS = RHS.
Hence proved.
(x) (1 + tan2 A1 + cot2 A) = (1 – tan A1 – cot A)2 = tan2 A
1 + tan2 A1 + cot2 A = sec2 Acosec2 A = sin2 Acos2 A = tan2 A
(1 – tan A1 – cot A)2 = (1 – tan A1 – 1/tan A)2 = (1 – tan A(tan A – 1)/tan A)2 = (–tan A)2 = tan2 A
All three expressions are equal.
Hence proved.
