CBSE NCERT Solutions For Class 10th Maths Chapter 8 : Introduction To Trigonometry. NCERT Solutins For Class 10 Mathematics. Exercise 8.1, Exercise 8.2, Exercise 8.3, Exercise 8.4.

---

NCERT Solutions for Class X Maths Chapter 8 Introduction To Trigonometry – Mathematics CBSE

Excercise 8.1

Page No: 181 

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

In Δ ABC,∠B = 90º
By Applying Pythagoras theorem , we get

AC² = AB² + BC² = (24)² + 7² = (576+49) cm² = 625 cm²
⇒ 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

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.

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

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.

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.

Answer

Let 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 = θ

A/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

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.

Excercise 8.2

Page No: 187

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

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,
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.

Excercise 8.3

Page No : 189

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 

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°

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

Page No : 193

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

(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.

(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

---

GO BACK TO CLASS X MATHS ALL CHAPTERS SOLUTION