Exercise 8.1
1. In Δ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine :
(i) sin A, cos A
(ii) sin C, cos C
Answer:
Let us draw a right angled triangle ABC, right angled at B.
Using Pythagoras theorem,
(i) 
(ii)
,
2. In adjoining figure, find tan P – cot R.
Answer:
3. If sin A =3/4, calculate cos A and tan A.
Answer:
Given: A triangle ABC in which
B =90
We know that sin A = BC/AC = 3/4
Let BC be 3k and AC will be 4k where k is a positive real number.
By Pythagoras theorem we get,
AC2 = AB2 + BC2
(4k)2 = AB2 + (3k)2
16k2 – 9k2 = AB2
AB2 = 7k2
AB = √7 k
cos A = AB/AC = √7 k/4k = √7/4
tan A = BC/AB = 3k/√7 k = 3/√7
4. Given 15 cot A = 8, find sin A and sec A.
Answer:
Let ΔABC be a right-angled triangle, right-angled at B.
We know that cot A = AB/BC = 8/15 (Given)
Let AB be 8k and BC will be 15k where k is a positive real number.
By Pythagoras theorem we get,
AC2 = AB2 + BC2
AC2 = (8k)2 + (15k)2
AC2 = 64k2 + 225k2
AC2 = 289k2
AC = 17 k
sin A = BC/AC = 15k/17k = 15/17
sec A = AC/AB = 17k/8 k = 17/8
5. Given sec θ = 13/12, calculate all other trigonometric ratios.
Answer:
Consider a triangle ABC in which
Let AB = 12k and AC = 13k
Then, using Pythagoras theorem,
(AC)2 = (AB)2 + (BC)2
(13 k)2 = (12 k)2 + (BC)2
169 k2 = 144 k2 + BC2
25 k2 = BC2
BC = 5 k
(I) Sin
= 
=
= 
(ii) Cos θ = 
=
= 
(iii)Tan θ = 
=
= 
(iv) Cot θ = 
=
= 
(v) Cosec θ = 
=
= 
6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Answer:
cosA=cos
But 
7. If cot θ =7/8, evaluate :
(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2θ
Answer:
Consider a triangle ABC
(ii)
8. If 3cot A = 4/3 , check whether (1-tan2A)/(1+tan2A) = cos2A – sin2A or not.
Answer:
Consider a triangle ABC AB=4cm, BC= 3cm
.
And 
9. In 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
Answer:
Consider a triangle ABC in which
.
(i)
(ii)
10. In Δ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Answer:
Given that, PR + QR = 25 , PQ = 5
Let PR be x. ∴ QR = 25 – x
By Pythagoras theorem ,
PR2 = PQ2 + QR2
x2 = (5)2 + (25 – x)2
x2 = 25 + 625 + x2 – 50x
50x = 650
x = 13
∴ PR = 13 cm
QR = (25 – 13) cm = 12 cm
sin P = QR/PR = 12/13
cos P = PQ/PR = 5/13
tan P = QR/PQ = 12/5
11. State whether the following are true or false. Justify your answer.
(i) The value of tan A is always less than 1.
(ii) sec A = 12/5 for some value of angle A.
(iii) cos A is the abbreviation used for the cosecant of angle A.
(iv) cot A is the product of cot and A.
(v) sin θ = 4/3 for some angle θ.
Answer:
i) False.
In ΔABC in which ∠B = 90º,
AB = 3, BC = 4 and AC = 5
Value of tan A = 4/3 which is greater than.
The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as
it will follow the Pythagoras theorem.
AC2 = AB2 + BC2
52 = 32 + 42
25 = 9 + 16
25 = 25
(ii) True.
Let a ΔABC in which ∠B = 90º,AC be 12k and AB be 5k, where k is a positive real number.
By Pythagoras theorem we get,
AC2 = AB2 + BC2
(12k)2 = (5k)2 + BC2
BC2 + 25k2 = 144k2
BC2 = 119k2
Such a triangle is possible as it will follow the Pythagoras theorem.
(iii) False.
Abbreviation used for cosecant of angle A is cosec A.cos A is the abbreviation used for cosine of angle A.
(iv) False.
cot A is not the product of cot and A. It is the cotangent of ∠A.
(v) False.
sin θ = Height/Hypotenuse
We know that in a right angled triangle, Hypotenuse is the longest side.
∴ sin θ will always less than 1 and it can never be 4/3 for any value of θ.
Exercise 8.2
1. Evaluate the following :
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan245° + cos230° – sin260°
(iii) cos 45°/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)
Answer:
(i) sin 60° cos 30° + sin 30° cos 60°
(ii) 2 tan245° + cos230° – sin260°
(iii) cos 45°/(sec 30° + cosec 30°)
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)
2. Choose the correct option and justify your choice :
(i) 2tan 30°/1+tan230° =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
(ii) 1-tan245°/1+tan245° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°/1-tan230° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
Answer:
(i) 2tan 30°/1+tan230° =
(A) sin 60° (B) cos 60° (C) tan 60° (D) sin 30°
= 
(ii) 1-tan245°/1+tan245° =
(A) tan 90° (B) 1 (C) sin 45° (D) 0
(iii) sin 2A = 2 sin A is true when A =
(A) 0° (B) 30° (C) 45° (D) 60°
(iv) 2tan30°/1-tan230° =
(A) cos 60° (B) sin 60° (C) tan 60° (D) sin 30°
3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.
Answer:
……….(i)
4. State whether the following are true or false. Justify your answer.
(i) sin (A + B) = sin A + sin B.
(ii) The value of sin θ increases as θ increases.
(iii) The value of cos θ increases as θ increases.
(iv) sin θ = cos θ for all values of θ.
(v) cot A is not defined for A = 0°.
Answer:
(i) False.
Let A = 30° and B = 60°, then
sin (A + B) = sin (30° + 60°) = sin 90° = 1 and,
sin A + sin B = sin 30° + sin 60°
= 1/2 + √3/2 = 1+√3/2
(ii) True.
sin 0° = 0
sin 30° = 1/2
sin 45° = 1/√2
sin 60° = √3/2
sin 90° = 1
Thus the value of sin θ increases as θ increases.
(iii) False.
cos 0° = 1
cos 30° = √3/2
cos 45° = 1/√2
cos 60° = 1/2
cos 90° = 0
Thus the value of cos θ decreases as θ increases.
(iv) True.
cot A = cos A/sin A
cot 0° = cos 0°/sin 0° = 1/0 = undefined.
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°
Answer:
(i) sin 18°/cos 72°
= sin (90° – 18°) /cos 72°
= cos 72° /cos 72° = 1
(ii) tan 26°/cot 64°
= tan (90° – 36°)/cot 64°
= cot 64°/cot 64° = 1
(iii) cos 48° – sin 42°
= cos (90° – 42°) – sin 42°
= sin 42° – sin 42° = 0
(iv) cosec 31° – sec 59°
= cosec (90° – 59°) – sec 59°
= 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
Answer:
(i) tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
= 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°
= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
= 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.
Answer:
tan 2A = cot (A- 18°)
⇒ cot (90° – 2A) = cot (A -18°)
Equating angles,
⇒ 90° – 2A = A- 18° ⇒ 108° = 3A
⇒ A = 36°
4. If tan A = cot B, prove that A + B = 90°.
Answer:
tan A = cot B
⇒ tan A = tan (90° – B)
⇒ A = 90° – B
⇒ A + B = 90°
5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Answer:
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°)
Equating angles,
90° – 4A= A- 20°
⇒ 110° = 5A
⇒ 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
Answer:
7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°.
Answer:
sin 67° + cos 75°
= sin (90° – 23°) + cos (90° – 15°)
= cos 23° + sin 15°
EXERCISE 8.4
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Answer:
3. Evaluate :
(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii) sin 25° cos 65° + cos 25° sin 65°
Answer:
(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii) sin 25° cos 65° + cos 25° sin 65°
Choose the correct option. Justify your choice.
4. (i) 9 sec2A – 9 tan2A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0 (B) 1 (C) 2 (D) – 1
(iii) (secA + tanA) (1 – sinA) =
(A) secA (B) sinA (C) cosecA (D) cosA
(iv) 1+tan2A/1+cot2A =
(A) sec2A (B) -1 (C) cot2A (D) tan2A
Answer:
(i) (i) 9 sec2A – 9 tan2A =
(A) 1 (B) 9 (C) 8 (D) 0
(ii) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
(A) 0 (B) 1 (C) 2 (D) – 1
(iii) (secA + tanA) (1 – sinA) =
(A) secA (B) sinA (C) cosecA (D) cosA
(iv) 1+tan2A/1+cot2A =
(A) sec2A (B) -1 (C) cot2A (D) tan2A
5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin2A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
(vi) √1 + sin A/1 – sin A = sec A+ tan A
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
Answer:
(i) (cosec θ – cot θ)2 = (1-cos θ)/(1+cos θ)
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
[Hint : Write the expression in terms of sin θ and cos θ]
(iv) (1 + sec A)/sec A = sin2A/(1-cos A)
[Hint : Simplify LHS and RHS separately]
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A
(vi) √1 + sin A/1 – sin A = sec A+ tan A
LHS = 1 + sin A/(1 – sin A) …..(1)
Multiplying and dividing by (1 + sin A)
⇒ (1 + sin A)(1 + sin A/1 – sin A)(1 + sin A)
= (1 + sin A)²/(1 – sin² A) [a² – b² = (a – b)(a + b)]
= (1 + sinA)/1 – sin² A
= 1 + sin A/cos² A
= 1 + sin A/cos A
= 1/cos A + sin A/cos A
= sec A + tan A
= R.H.S
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
[Hint : Simplify LHS and RHS separately]
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A