In This Post we are providing Chapter 8 Introduction to Trigonometry NCERT Solutions for Class 10 Maths which will be beneficial for students. These solutions are updated according to 2021-22 syllabus. These Introduction to Trigonometry Class 10 solutions can be really helpful in the preparation of Board exams and will provide you with in depth detail of the chapter.
We have solved every question stepwise so you don’t have to face difficulty in understanding the solutions. It will also give you concepts that are important for overall development of students. Class 10 Maths Introduction to Trigonometry NCERT Written Solutions & Video Solution will be useful in higher classes as well because variety of questions related to these concepts can be asked so you must study and understand them properly.
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
(i) sin A, cos A
(ii) sin C, cos C
Answer
In Δ ABC,∠B = 90º
By Applying Pythagoras theorem , we get
AC2 = AB2 + BC2 = (24)2 + 72 = (576+49) cm2 = 625 cm2
⇒ AC = 25
In Δ ABC,∠B = 90º
By Applying Pythagoras theorem , we get
⇒ AC = 25
(i) sin A = BC/AC = 7/25 cos A = AB/AC = 24/25
(ii) sin C = AB/AC = 24/25
cos C = BC/AC = 7/25
2. In Fig. 8.13, find tan P – cot R.
Answer
By Applying Pythagoras theorem in ΔPQR , we get
PR2 = PQ2 + QR2 = (13)2 = (12)2 + QR2 = 169 = 144 + QR2
⇒ QR2 = 25 ⇒ QR = 5 cm
(ii) sin C = AB/AC = 24/25
cos C = BC/AC = 7/25
2. In Fig. 8.13, find tan P – cot R.
Answer
PR2 = PQ2 + QR2 = (13)2 = (12)2 + QR2 = 169 = 144 + QR2
⇒ QR2 = 25 ⇒ QR = 5 cm
Now,
tan P = QR/PQ = 5/12
cot R = QR/PQ = 5/12
A/q
tan P – cot R = 5/12 – 5/12 = 0
3. If sin A =3/4, calculate cos A and tan A.
Answer
Let ΔABC be a right-angled triangle, right-angled at B.
tan P = QR/PQ = 5/12
cot R = QR/PQ = 5/12
A/q
tan P – cot R = 5/12 – 5/12 = 0
3. If sin A =3/4, calculate cos A and tan A.
Answer
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.
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
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
Let ΔABC be a right-angled triangle, right-angled at B.
We know that sec θ = OP/OM = 13/12 (Given)
Let OP be 13k and OM will be 12k where k is a positive real number.
Let OP be 13k and OM will be 12k where k is a positive real number.
By Pythagoras theorem we get,OP2 = OM2 + MP2
(13k)2 = (12k)2 + MP2
169k2 – 144k2 = MP2
MP2 = 25k2
MP = 5
Now,
sin θ = MP/OP = 5k/13k = 5/13
cos θ = OM/OP = 12k/13k = 12/13
tan θ = MP/OM = 5k/12k = 5/12
cot θ = OM/MP = 12k/5k = 12/5
cosec θ = OP/MP = 13k/5k = 13/5
6. If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
AnswerLet ΔABC in which CD ⊥ AB.
A/q,
cos A = cos B
⇒ AD/AC = BD/BC
⇒ AD/BD = AC/BC
Let AD/BD = AC/BC = k
⇒ AD = kBD …. (i)
⇒ AC = kBC …. (ii)
By applying Pythagoras theorem in ΔCAD and ΔCBD we get,
CD2 = AC2 – AD2 …. (iii)
and also CD2 = BC2 – BD2 …. (iv)
From equations (iii) and (iv) we get,
AC2 – AD2 = BC2 – BD2
⇒ (kBC)2 – (k BD)2 = BC2 – BD2
⇒ k2 (BC2 – BD2) = BC2 – BD2
⇒ k2 = 1
⇒ k = 1
Putting this value in equation (ii), we obtain
AC = BC
⇒ ∠A = ∠B (Angles opposite to equal sides of a triangle are equal-isosceles triangle)
7. If cot θ =7/8, evaluate :
(i)(1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ)
(ii) cot2θ
Answer
Let ΔABC in which ∠B = 90º and ∠C = θA/q,
cot θ = BC/AB = 7/8
Let BC = 7k and AB = 8k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC2
AC2 = (8k)2 + (7k)2
AC2 = 64k2 + 49k2
AC2 = 113k2
AC = √113 k
sin θ = AB/AC = 8k/√113 k = 8/√113
and cos θ = BC/AC = 7k/√113 k = 7/√113
(i) (1+sin θ )(1-sin θ)/(1+cos θ)(1-cos θ) = (1-sin2θ)/(1-cos2θ) = {1 – (8/√113)2}/{1 – (7/√113)2}
= {1 – (64/113)}/{1 – (49/113)} = {(113 – 64)/113}/{(113 – 49)/113} = 49/64
(ii) cot2θ = (7/8)2 = 49/64
8. If 3cot A = 4/3 , check whether (1-tan2A)/(1+tan2A) = cos2A – sin2A or not.
Answer
Let ΔABC in which ∠B = 90º,
A/q,
cot A = AB/BC = 4/3
Let AB = 4k and BC = 3k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC2
AC2 = (4k)2 + (3k)2
AC2 = 16k2 + 9k2
AC2 = 25k2
AC = 5k
tan A = BC/AB = 3/4
sin A = BC/AC = 3/5
cos A = AB/AC = 4/5
L.H.S. = (1-tan2A)/(1+tan2A) = 1- (3/4)2/1+ (3/4)2 = (1- 9/16)/(1+ 9/16) = (16-9)/(16+9) = 7/25
R.H.S. = cos2A – sin2A = (4/5)2 – (3/4)2 = (16/25) – (9/25) = 7/25
R.H.S. = L.H.S.
Hence, (1-tan2A)/(1+tan2A) = cos2A – sin2A
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
Let ΔABC in which ∠B = 90º,
A/q,
tan A = BC/AB = 1/√3
Let AB = √3 k and BC = k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC2
AC2 = (√3 k)2 + (k)2
AC2 = 3k2 + k2
AC2 = 4k2
AC = 2k
sin A = BC/AC = 1/2 cos A = AB/AC = √3/2 ,
sin C = AB/AC = √3/2 cos A = BC/AC = 1/2
(i) sin A cos C + cos A sin C = (1/2×1/2) + (√3/2×√3/2) = 1/4+3/4 = 4/4 = 1
(ii) cos A cos C – sin A sin C = (√3/2×1/2) – (1/2×√3/2) = √3/4 – √3/4 = 0
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
Answer
Let ΔABC in which ∠B = 90º,
A/q,
cot A = AB/BC = 4/3
Let AB = 4k and BC = 3k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC2
AC2 = (4k)2 + (3k)2
AC2 = 16k2 + 9k2
AC2 = 25k2
AC = 5k
tan A = BC/AB = 3/4
sin A = BC/AC = 3/5
cos A = AB/AC = 4/5
L.H.S. = (1-tan2A)/(1+tan2A) = 1- (3/4)2/1+ (3/4)2 = (1- 9/16)/(1+ 9/16) = (16-9)/(16+9) = 7/25
R.H.S. = cos2A – sin2A = (4/5)2 – (3/4)2 = (16/25) – (9/25) = 7/25
R.H.S. = L.H.S.
Hence, (1-tan2A)/(1+tan2A) = cos2A – sin2A
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
Let ΔABC in which ∠B = 90º,
A/q,
tan A = BC/AB = 1/√3Let AB = √3 k and BC = k, where k is a positive real number.
By Pythagoras theorem in ΔABC we get.
AC2 = AB2 + BC2
AC2 = (√3 k)2 + (k)2
AC2 = 3k2 + k2
AC2 = 4k2
AC = 2k
sin A = BC/AC = 1/2 cos A = AB/AC = √3/2 ,
sin C = AB/AC = √3/2 cos A = BC/AC = 1/2
(i) sin A cos C + cos A sin C = (1/2×1/2) + (√3/2×√3/2) = 1/4+3/4 = 4/4 = 1
(ii) cos A cos C – sin A sin C = (√3/2×1/2) – (1/2×√3/2) = √3/4 – √3/4 = 0
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.
(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.
(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°
= (√3/2×√3/2) + (1/2×1/2) = 3/4 + 1/4 = 4/4 = 1
(ii) 2 tan245° + cos230° – sin260°
= 2×(1)2 + (√3/2)2 – (√3/2)2 = 2
(iii) cos 45°/(sec 30° + cosec 30°)
= 1/√2/(2/√3 + 2) = 1/√2/{(2+2√3)/√3)
= √3/√2×(2+2√3) = √3/(2√2+2√6)
= √3(2√6-2√2)/(2√6+2√2)(2√6-2√2)
= 2√3(√6-√2)/(2√6)2-(2√2)2
= 2√3(√6-√2)/(24-8) = 2√3(√6-√2)/16
= √3(√6-√2)/8 = (√18-√6)/8 = (3√2-√6)/8
(iv) (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)
= (1/2+1-2/√3)/(2/√3+1/2+1)
= (3/2-2/√3)/(3/2+2/√3)
= (3√3-4/2√3)/(3√3+4/2√3)
= (3√3-4)/(3√3+4)
= (3√3-4)(3√3-4)/(3√3+4)(3√3-4)
= (3√3-4)2/(3√3)2-(4)2
= (27+16-24√3)/(27-16)
= (43-24√3)/11]
(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)
= 5(1/2)2+4(2/√3)2-12/(1/2)2+(√3/2)2
= (5/4+16/3-1)/(1/4+3/4)
= (15+64-12)/12/(4/4)
= 67/12
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) (A) is correct.
2tan 30°/1+tan230° = 2(1/√3)/1+(1/√3)2
= (2/√3)/(1+1/3) = (2/√3)/(4/3)
= 6/4√3 = √3/2 = sin 60°
(ii) (D) is correct.
1-tan245°/1+tan245° = (1-12)/(1+12)
= 0/2 = 0
(iii) (A) is correct.
sin 2A = 2 sin A is true when A =
= As sin 2A = sin 0° = 0
2 sin A = 2sin 0° = 2×0 = 0
2 sin A = 2sin 0° = 2×0 = 0
or,
sin 2A = 2sin A cos A
⇒2sin A cos A = 2 sin A⇒ 2cos A = 2 ⇒ cos A = 1
⇒ A = 0°
(iv) (C) is correct.
2tan30°/1-tan230° = 2(1/√3)/1-(1/√3)2
= (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°
3. If tan (A + B) = √3 and tan (A – B) = 1/√3; 0° < A + B ≤ 90°; A > B, find A and B.
Answer
tan (A + B) = √3
⇒ tan (A + B) = tan 60°⇒ (A + B) = 60° … (i)
tan (A – B) = 1/√3
⇒ tan (A – B) = tan 30°⇒ (A – B) = 30° … (ii)
Adding (i) and (ii), we get
A + B + A – B = 60° + 30°
2A = 90°
A= 45°
Putting the value of A in equation (i)
45° + B = 60°
⇒ B = 60° – 45°
⇒ B = 15°
Thus, A = 45° and B = 15°
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,
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°
(ii) True.
(iii) False.
(iv) True.
= 1/2 + √3/2 = 1+√3/2
(ii) True.
sin 0° = 0
sin 30° = 1/2
sin 45° = 1/√2
sin 60° = √3/2sin 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° = 1/2cos 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.
Excercise 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
= 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
A/q,
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
A/q,
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
A/q,
sec 4A = cosec (A – 20°)
⇒ cosec (90° – 4A) = cosec (A – 20°)
Equating angles,
90° – 4A= A- 20°
⇒ 110° = 5A
⇒ A = 22°
90° – 4A= A- 20°
⇒ 110° = 5A
⇒ A = 22°
Page No : 190
6. If A, B and C are interior angles of a triangle ABC, then show that
sin (B+C/2) = cos A/2
Answer
In a triangle, sum of all the interior angles
A + B + C = 180°
⇒ B + C = 180° – A
⇒ (B+C)/2 = (180°-A)/2
⇒ (B+C)/2 = (90°-A/2)
⇒ sin (B+C)/2 = sin (90°-A/2)
⇒ sin (B+C)/2 = cos A/2
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°
Excercise 8.4
1. Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.
Answer
cosec2A – cot2A = 1
⇒ cosec2A = 1 + cot2A
⇒ 1/sin2A = 1 + cot2A
⇒sin2A = 1/(1+cot2A)
Now,
sin2A = 1/(1+cot2A)
⇒ 1 – cos2A = 1/(1+cot2A)
⇒cos2A = 1 – 1/(1+cot2A)
⇒cos2A = (1-1+cot2A)/(1+cot2A)
⇒ 1/sec2A = cot2A/(1+cot2A)
⇒ secA = (1+cot2A)/cot2A
also,
tan A = sin A/cos A and cot A = cos A/sin A
⇒ tan A = 1/cot A
2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Answer
We know that,
sec A = 1/cos A
⇒ cos A = 1/sec A
also,
cos2A + sin2A = 1
⇒ sin2A = 1 – cos2A
⇒ sin2A = 1 – (1/sec2A)
⇒ sin2A = (sec2A-1)/sec2A
also,
sin A = 1/cosec A
⇒cosec A = 1/sin A
Now,
sec2A – tan2A = 1
⇒ tan2A = sec2A + 1
also,
tan A = 1/cot A
⇒ cot A = 1/tan A
3. Evaluate :
(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii) sin 25° cos 65° + cos 25° sin 65°
Answer
(i) (sin263° + sin227°)/(cos217° + cos273°)
= [sin2(90°-27°) + sin227°]/[cos2(90°-73°) + cos273°)]
= (cos227° + sin227°)/(sin227° + cos273°)
= 1/1 =1 (∵ sin2A + cos2A = 1)
(ii) sin 25° cos 65° + cos 25° sin 65°
= sin(90°-25°) cos 65° + cos(90°-65°) sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= cos265° + sin265° = 1
4. Choose the correct option. Justify your choice.
(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
⇒ cosec2A = 1 + cot2A
⇒ 1/sin2A = 1 + cot2A
⇒sin2A = 1/(1+cot2A)
Now,
sin2A = 1/(1+cot2A)
⇒ 1 – cos2A = 1/(1+cot2A)
⇒cos2A = 1 – 1/(1+cot2A)
⇒cos2A = (1-1+cot2A)/(1+cot2A)
⇒ 1/sec2A = cot2A/(1+cot2A)
⇒ secA = (1+cot2A)/cot2A
also,
tan A = sin A/cos A and cot A = cos A/sin A
⇒ tan A = 1/cot A
2. Write all the other trigonometric ratios of ∠A in terms of sec A.
Answer
We know that,
sec A = 1/cos A
⇒ cos A = 1/sec A
also,
cos2A + sin2A = 1
⇒ sin2A = 1 – cos2A
⇒ sin2A = 1 – (1/sec2A)
⇒ sin2A = (sec2A-1)/sec2A
also,
sin A = 1/cosec A
⇒cosec A = 1/sin A
Now,
sec2A – tan2A = 1
⇒ tan2A = sec2A + 1
also,
tan A = 1/cot A
⇒ cot A = 1/tan A
3. Evaluate :
(i) (sin263° + sin227°)/(cos217° + cos273°)
(ii) sin 25° cos 65° + cos 25° sin 65°
Answer
(i) (sin263° + sin227°)/(cos217° + cos273°)
= [sin2(90°-27°) + sin227°]/[cos2(90°-73°) + cos273°)]
= (cos227° + sin227°)/(sin227° + cos273°)
= 1/1 =1 (∵ sin2A + cos2A = 1)
(ii) sin 25° cos 65° + cos 25° sin 65°
= sin(90°-25°) cos 65° + cos(90°-65°) sin 65°
= cos 65° cos 65° + sin 65° sin 65°
= cos265° + sin265° = 1
4. Choose the correct option. Justify your choice.
(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) (B) is correct.
9 sec2A – 9 tan2A
= 9 (sec2A – tan2A)
= 9×1 = 9 (∵ sec2 A – tan2 A = 1)
(ii) (C) is correct
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
= (1 + sin θ/cos θ + 1/cos θ) (1 + cos θ/sin θ – 1/sin θ)
= (cos θ+sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ
= (cos θ+sin θ)2-12/(cos θ sin θ)
= (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ)
= (1+ 2cos θ sin θ -1)/(cos θ sin θ)
= (2cos θ sin θ)/(cos θ sin θ) = 2
(iii) (D) is correct.
(secA + tanA) (1 – sinA)
= (1/cos A + sin A/cos A) (1 – sinA)
= (1+sin A/cos A) (1 – sinA)
= (1 – sin2A)/cos A
= cos2A/cos A = cos A
(iv) (D) is correct.
1+tan2A/1+cot2A
= (1+1/cot2A)/1+cot2A
= (cot2A+1/cot2A)×(1/1+cot2A)
= 1/cot2A = tan2A
5. Prove the following identities, where the angles involved are acute angles for which the
expressions are defined.
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.
(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)
L.H.S. = (cosec θ – cot θ)2
= (cosec2θ + cot2θ – 2cosec θ cot θ)
= (1/sin2θ + cos2θ/sin2θ – 2cos θ/sin2θ)
= (1 + cos2θ – 2cos θ)/(1 – cos2θ)
= (1-cos θ)2/(1 – cosθ)(1+cos θ)
= (1-cos θ)/(1+cos θ) = R.H.S.
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
= [cos2A + (1+sin A)2]/(1+sin A)cos A
= (cos2A + sin2A + 1 + 2sin A)/(1+sin A)cos A
= (1 + 1 + 2sin A)/(1+sin A)cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A = 2 sec A = R.H.S.
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]
= sin2θ/[cos θ(sin θ-cos θ)] + cos2θ/[sin θ(cos θ-sin θ)]
= sin2θ/[cos θ(sin θ-cos θ)] – cos2θ/[sin θ(sin θ-cos θ)]
= 1/(sin θ-cos θ) [(sin2θ/cos θ) – (cos2θ/sin θ)]
= 1/(sin θ-cos θ) × [(sin3θ – cos3θ)/sin θ cos θ]
= [(sin θ-cos θ)(sin2θ+cos2θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
= (1 + sin θ cos θ)/sin θ cos θ
= 1/sin θ cos θ + 1
= 1 + sec θ cosec θ = R.H.S.
(iv) (1 + sec A)/sec A = sin2A/(1-cos A)
L.H.S. = (1 + sec A)/sec A
= (1 + 1/cos A)/1/cos A
= (cos A + 1)/cos A/1/cos A
= cos A + 1
R.H.S. = sin2A/(1-cos A)
= (1 – cos2A)/(1-cos A)
= (1-cos A)(1+cos A)/(1-cos A)
= cos A + 1
L.H.S. = R.H.S.
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
Dividing Numerator and Denominator by sin A,
= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)
= (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A – cot2A = 1)
= [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A)
= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)
= (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
= cot A + cosec A = R.H.S.
Dividing Numerator and Denominator of L.H.S. by cos A,
= sec A + tan A = R.H.S.
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
L.H.S. = (sin θ – 2sin3θ)/(2cos3θ – cos θ)
= [sin θ(1 – 2sin2θ)]/[cos θ(2cos2θ- 1)]
= sin θ[1 – 2(1-cos2θ)]/[cos θ(2cos2θ -1)]
= [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)]
= tan θ = R.H.S.
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= (sin2A + cosec2A + 2 sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)
= (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A
= 1 + 2 + 2 + 2 + tan2A + cot2A
= 7+tan2A+cot2A = R.H.S.
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
= (1/sin A – sin A)(1/cos A – cos A)
= [(1-sin2A)/sin A][(1-cos2A)/cos A]
= (cos2A/sin A)×(sin2A/cos A)
= cos A sin A
R.H.S. = 1/(tan A+cotA)
= 1/(sin A/cos A +cos A/sin A)
= 1/[(sin2A+cos2A)/sin A cos A]
= cos A sin A
L.H.S. = R.H.S.
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
L.H.S. = (1+tan2A/1+cot2A)
= (1+tan2A/1+1/tan2A)
= 1+tan2A/[(1+tan2A)/tan2A]
= tan2A
(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)
L.H.S. = (cosec θ – cot θ)2
= (cosec2θ + cot2θ – 2cosec θ cot θ)
= (1/sin2θ + cos2θ/sin2θ – 2cos θ/sin2θ)
= (1 + cos2θ – 2cos θ)/(1 – cos2θ)
= (1-cos θ)2/(1 – cosθ)(1+cos θ)
= (1-cos θ)/(1+cos θ) = R.H.S.
(ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
= [cos2A + (1+sin A)2]/(1+sin A)cos A
= (cos2A + sin2A + 1 + 2sin A)/(1+sin A)cos A
= (1 + 1 + 2sin A)/(1+sin A)cos A
= (2+ 2sin A)/(1+sin A)cos A
= 2(1+sin A)/(1+sin A)cos A
= 2/cos A = 2 sec A = R.H.S.
(iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
= [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
= [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cos θ-sin θ)/cos θ]
= sin2θ/[cos θ(sin θ-cos θ)] + cos2θ/[sin θ(cos θ-sin θ)]
= sin2θ/[cos θ(sin θ-cos θ)] – cos2θ/[sin θ(sin θ-cos θ)]
= 1/(sin θ-cos θ) [(sin2θ/cos θ) – (cos2θ/sin θ)]
= 1/(sin θ-cos θ) × [(sin3θ – cos3θ)/sin θ cos θ]
= [(sin θ-cos θ)(sin2θ+cos2θ+sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
= (1 + sin θ cos θ)/sin θ cos θ
= 1/sin θ cos θ + 1
= 1 + sec θ cosec θ = R.H.S.
(iv) (1 + sec A)/sec A = sin2A/(1-cos A)
L.H.S. = (1 + sec A)/sec A
= (1 + 1/cos A)/1/cos A
= (cos A + 1)/cos A/1/cos A
= cos A + 1
R.H.S. = sin2A/(1-cos A)
= (1 – cos2A)/(1-cos A)
= (1-cos A)(1+cos A)/(1-cos A)
= cos A + 1
L.H.S. = R.H.S.
(v) (cos A–sin A+1)/(cos A+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
L.H.S. = (cos A–sin A+1)/(cos A+sin A–1)
Dividing Numerator and Denominator by sin A,
= (cos A–sin A+1)/sin A/(cos A+sin A–1)/sin A
= (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)
= (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2A – cot2A = 1)
= [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A)
= [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)
= (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
= cot A + cosec A = R.H.S.
Dividing Numerator and Denominator of L.H.S. by cos A,
= sec A + tan A = R.H.S.
(vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
L.H.S. = (sin θ – 2sin3θ)/(2cos3θ – cos θ)
= [sin θ(1 – 2sin2θ)]/[cos θ(2cos2θ- 1)]
= sin θ[1 – 2(1-cos2θ)]/[cos θ(2cos2θ -1)]
= [sin θ(2cos2θ -1)]/[cos θ(2cos2θ -1)]
= tan θ = R.H.S.
(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7+tan2A+cot2A
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= (sin2A + cosec2A + 2 sin A cosec A) + (cos2A + sec2A + 2 cos A sec A)
= (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A
= 1 + 2 + 2 + 2 + tan2A + cot2A
= 7+tan2A+cot2A = R.H.S.
(ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
L.H.S. = (cosec A – sin A)(sec A – cos A)
= (1/sin A – sin A)(1/cos A – cos A)
= [(1-sin2A)/sin A][(1-cos2A)/cos A]
= (cos2A/sin A)×(sin2A/cos A)
= cos A sin A
R.H.S. = 1/(tan A+cotA)
= 1/(sin A/cos A +cos A/sin A)
= 1/[(sin2A+cos2A)/sin A cos A]
= cos A sin A
L.H.S. = R.H.S.
(x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A
L.H.S. = (1+tan2A/1+cot2A)
= (1+tan2A/1+1/tan2A)
= 1+tan2A/[(1+tan2A)/tan2A]
= tan2A