# NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry – This article includes detailed NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry, designed and reviewed by subject experts at GFG.

NCERT Class 10 Maths Chapter 8 Introduction to Trigonometry comprises the following topics:

### Class 10 Maths NCERT Solutions Chapter 8 Introduction to Trigonometry Exercises

This article provides solutions to all the problems asked in Class 10 Maths Chapter 8 Introduction to Trigonometry of your NCERT textbook in a step-by-step manner. Solutions to all the exercises in the NCERT Class 9 Maths Chapter 7 Triangles are regularly revised to check errors and updated according to the latest CBSE Syllabus 2023-24 and guidelines.

The solutions to all the ercercises in NCERT Chapter 8 Introduction to Trigonometry have been collectively covered in NCERT Solutions for Class 10 Maths.

## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry: Exercise 8.1

### (i) sin A, cos A (ii) sin C, cos C

Solution:

Using Pythagoras theorem for Î”ABC

AC2 = AB2 + BC2

= (24 cm)2 + (7 cm)2

= (576 + 49) cm2

= 625 cm2

âˆ´AC = 25 cm

(i) sin A = opp/hyp

sin A = 7/25

cos A = adj/hyp = 24/25

cos A = 24/25

(ii) sin C = opp/hyp

sin C = 24/25

cos C = 7/25

### Question 2. In Fig., find tan P â€“ cot R.

Solution:

Applying Pythagoras theorem for Î”PQR, we obtain

PR2 = PQ2 + QR2

(13 cm)2 = (12 cm)2 + QR2

169 cm2 = 144 cm2 + QR2

25 cm2 = QR2

QR = 5 cm

tan P = 5/12

cot R = 5/12

tan P â€“ cot R = 5/12 â€“ 5/12 = 0

### Question 3. If sin A = 3/4, calculate cos A and tan A.

Solution:

Using sin2A + cos2A = 1

(3/4)2 + cos2A = 1

cos2A = 1 â€“ (3/4)2 = 1 â€“ 9/16

cos A = 71/2/4

tan A = sin A/cos A

tan A = (3/4)/(71/2/4)

tan A = 3/71/2

### Question 4: Given 15 cot A = 8. Find sin A and sec A

Solution:

Given, 15 cot A = 8

cot A = 8/15

tan A = 1/cot A

tan A = 15/8

Using, 1 + tan2A = sec2A

1 + (15/8)= sec2A

289/64 = sec2A

sec A = 17/8

We know, cos2A = 1/sec2A

cos2A = 64/289

sin2A = 1 â€“ cos2A

sin2A = 225/289

sin A = 15/17

### Question 5: Given sec Î¸ = 13/12, calculate all other trigonometric ratios.

Solution.

Using Pythagoras theorem,

sin Î¸ = 5/13

cos Î¸ = 12/13

tan Î¸ = 5/12

cosec Î¸ = 13/5

cot Î¸ = 12/5

### Question 6: If âˆ A and âˆ B are acute angles such that cos A = cos B, then show that âˆ A = âˆ B.

Solution:

Let us consider a Î”ABC in which CD âŠ¥ AB.

It is given that cos A = cos B

We need to prove âˆ A = âˆ B. To prove this, we need to extend AC to P such that BC = CP.

From equation (1), we obtain

AD/BD = AC/CP    (BC = CP by construction)

By using the converse of B.P.T (Basic Proportionality Theorem),

CD||BP

âˆ ACD = âˆ CPB (Corresponding angles) â€¦ (3)

And, âˆ BCD = âˆ CBP (Alternate interior angles) â€¦ (4)

By construction, we have BC = CP.

âˆ´ âˆ CBP = âˆ CPB (Angle opposite to equal sides of a triangle) â€¦ (5)

From equations (3), (4), and (5), we obtain

âˆ ACD = âˆ BCD â€¦ (6)

âˆ ACD = âˆ BCD (Using equation (6))

âˆ CDA = âˆ CDB (Both 90Â°)

Therefore, the remaining angles should be equal.

â‡’ âˆ A = âˆ B

### (ii) cot2Î¸

Solution:

(i) Using (a + b) * (a â€“ b) = a2 â€“ b2 in numerator and denominator

We get

(1 â€“ sin2Î¸)/(1 â€“ cos2Î¸)

Using sin2Î¸ + cos2Î¸ = 1

We get

cos2Î¸/sin2Î¸ = cot2Î¸

Now

cot2Î¸ = (7/8)2 = 49/64

(ii) cot2Î¸ = (7/8)2 = 49/64

### Question 8. If 3 cot A = 4, Check whether (1 â€“ tan2A)/(1 + tan2A) = cos2A â€“ sin2A

Solution.

We know that, tanA = sinA / cosA   â€¦.(1)

Using (1) on L.H.S

= (1 â€“ sin2A/cos2A)/(1 + sin2A/cos2A)

which on rearranging becomes

= (cos2A â€“ sin2A)/(cos2A + sin2A)

Using the identity,

cos2A + sin2A = 1

LHS becomes

= (cos2A â€“ sin2A)

This is equal to RHS.

LHS = RHS (for every value of cot A)

Hence, Proved.

### (ii) cos A cos C âˆ’ sin A sin C

Solution:

Using Pythagoras theorem

(AB)2 + (BC)2 = (AC)2

(31/2)2 + (1)2 = (AC)2

which gives

AC = 2 cm

Using formulas

sin A = 1/2

sin C = 31/2/2

cos A = 31/2/2

cos C = 1/2

Now, (i) sin A cos C + cos A sin C

Substituting the values

= (1/2) * (1/2) + (31/2/2) * (31/2/2)

= 1/4 + 3/4

= 1

Now, (ii) cos A cos C âˆ’ sin A sin C

Substituting the values

= (31/2/2) * (1/2) â€“ (1/2) * (31/2/2)

= 31/2/4 â€“ 31/2/4

= 0

### Question 10: In Î”PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P, and tan P.

Solution:

Given that, PR + QR = 25

PQ = 5

Let PR be x cm.

Therefore, QR = 25 âˆ’ x cm

Applying Pythagoras theorem in Î”PQR, we obtain

PR2 = PQ2 + QR2

x2 = (5)2 + (25 âˆ’ x)2

x2 = 25 + 625 + x2 âˆ’ 50x

50x = 650

x = 13

Therefore, PR = 13 cm

QR = (25 âˆ’ 13) cm = 12 cm

Now,

sin P = QR/PR = 12/13

cos P = PQ/PR = 5/13

tan P = QR/PQ = 12/5

### (v) sin Î¸ = 4/3, for some angle Î¸

Solution:

(i) Consider a Î”PQR, right-angled at Q as shown below.

Here tan P = 12/5 which is surely greater than 1.

Therefore, the statement is false.

(ii) Consider Î”ABC with AB = 5 cm, AC = 12 cm and BC = x cm

Using Pythagoras theorem in Î”ABC

(AB)2 + (BC)2 = (AC)2

52 + x2 = 122

x = (144 â€“ 25)1/2

x = (119)1/2

x = 10.9 cm

AB < BC < AC

So this triangle is valid,

Therefore, given statement is true.

(iii) Abbreviation used for cosecant A is cosec A. And cos A is the abbreviation used for cosine A.

Hence, the given statement is false.

(iv) cot A is not the product of cot and A. It is the cotangent of âˆ A.

Hence, the given statement is false.

(v) sin Î¸ = 4/3

In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin Î¸ is not possible.

Hence, the given statement is false

## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry: Exercise 8.2

### (i) sin 60Â° cos 30Â° + sin 30Â° cos 60Â°

Solution:

Formulas to be used : sin 30Â° = 1/2

cos 30Â° = âˆš3/2

sin 60Â° = 3/2

cos 60Â° = 1/2

=> (âˆš3/2) * (âˆš3/2) + (1/2) * (1/2)

=> 3/4 +1/4

=> 4 /4

=> 1

### (ii) 2 tan245Â° + cos230Â° â€“ sin260Â°

Solution:

Formulas to be used : sin 60Â° = âˆš3/2

cos 30Â° = âˆš3/2

tan 45Â° = 1

=> 2(1)(1) + (âˆš3/2)(âˆš3/2)-(âˆš3/2)(âˆš3/2)

=> 2 + 3/4 – 3/4

=> 2

### (iii) cos 45Â°/(sec 30Â°+cosec 30Â°)

Solution:

Formulas to be used : cos 45Â° = 1/âˆš2

sec 30Â° = 2/âˆš3

cosec 30Â° = 2

=> 1/âˆš2 / (2/âˆš3 + 2)

=> 1/âˆš2 / (2+2âˆš3)/âˆš3

=> âˆš3/âˆš2Ã—(2+2 âˆš3) = âˆš3/(2âˆš2+2âˆš6)

=> âˆš3(2âˆš6-2âˆš2)/(2âˆš6+2âˆš2)(2âˆš6-2âˆš2)

=> 2âˆš3(âˆš6-âˆš2) / (2âˆš6)Â²-(2âˆš2)Â²

=> 2âˆš3(âˆš6-âˆš2)/(24-8) = 2 âˆš3(âˆš6-âˆš2)/16

=> âˆš3(âˆš6-âˆš2)/8

=> (âˆš18-âˆš6)/8

=> (3âˆš2-âˆš6)/8

### (iv) (sin 30Â° + tan 45Âº – cosec 60Â°)/(sec 30Â° + cos 60Â° + cot 45Â°)

Solution:

Formulas to be used : sin 30Â° = 1/2

tan 45Â° = 1

cosec 60Â° = 2/âˆš3

sec 30Â° = 2/âˆš3

cos 60Â° = 1/2

cot 45Â° = 1

=> (1/2+1-2/âˆš3) / (2/âˆš3+1/2+1)

=> (3/2-2/âˆš3)/(3/2+2/âˆš3)

=> (3âˆš3-4/2 âˆš3)/(3âˆš3+4/2 âˆš3)

=> (3âˆš3-4)(3âˆš3-4)/(3âˆš3+4)(3âˆš3-4)

=> (27+16-24âˆš3) / (27-16)

=> (43-24âˆš3)/11

### (v) (5cos260Â° + 4sec2 30Â° – tan245Â°)/(sin230Â° + cosÂ²30Â°)

Solution:

Formulas to be used : cos 60Â° = 1/2

sec 30Â° = 2/âˆš3

tan 45Â° = 1

sin 30Â° = 1/2

cos 30Â° = âˆš3/2

=> 5(1/2)2+4(2/âˆš3)Â²-1Â²/(1/2)+(âˆš3/2)

=> (5/4+16/3-1) / (1/4+3/4)

=> (15+64-12) / 12/(4/4)

=> 67/12

### Question 2. Choose the correct option and justify your choice:

(i) 2tan 30Â°/1+tan230Â° =

(A) sin 60Â°            (B) cos 60Â°          (C) tan 60Â°            (D) sin 30Â°

(ii) 1-tan245Â°/1+tan245Â° =

(A) tan 90Â°            (B) 1                    (C) sin 45Â°            (D) 0

(iii)  sin 2A = 2 sin A is true when A =

(A) 0Â°                   (B) 30Â°                  (C) 45Â°                 (D) 60Â°

(iv) 2tan30Â°/1-tan230Â° =

(A) cos 60Â°          (B) sin 60Â°             (C) tan 60Â°           (D) sin 30Â°

Solution:

(i) In the given equation, substituting the value of tan 30Â°

As tan 30Â° = 1/âˆš3

2tan 30Â°/1+tan230Â° = 2(1/âˆš3)/1+(1/âˆš3)2

=> (2/âˆš3)/(1+1/3) = (2/âˆš3)/(4/3)

=> 6/4âˆš3 = âˆš3/2

=> sin 60Â°

The ans is sin 60Â°.

The correct option is (A).

(ii) In the given equation, substituting the of tan 45Â°

As tan 45Â° = 1

1-tan245Â°/1+tan245Â° = (1-12)/(1+12)

= 0/2 => 0

The ans is 0.

The correct option is (D).

(iii) sin 2A = 2 sin A is true when A = 0Â°

sin 2A = sin 0Â° = 0

2 sin A = 2 sin 0Â° = 2 Ã— 0 = 0

Another way :

sin 2A = 2sin A cos A

=> 2sin A cos A = 2 sin A

=> 2cos A = 2 => cos A = 1

Now, we have to check which degree value has to be applied, to get the solution as 1.

When 0 degree is applied to cos value we get 1, i.e., cos 0 = 1

Hence, A = 0Â°

The correct option is (A).

(iv) As tan 30Â° = 1/âˆš3

2tan30Â°/1-tan230Â° =  2(1/âˆš3)/1-(1/âˆš3)2

=> (2/âˆš3)/(1-1/3) = (2/âˆš3)/(2/3) = âˆš3 = tan 60Â°

The correct option is (C).

### Question 3. If tan (A + B) = âˆš3 and tan (A â€“ B) = 1/âˆš3, 0Â° < A + B â‰¤ 90Â°; A > B, find A and B.

Solution:

tan (A + B) = âˆš3

tan (A + B) = tan 60Â°

(A + B) = 60Â° â€¦ (i)

tan (A â€“ B) = 1/âˆš3

tan (A â€“ B) = tan 30Â°

(A â€“ B) = 30Â° â€¦  (ii)

Now add the equation (i) and (ii), we get

A + B + A â€“ B = 60Â° + 30Â°

A= 45Â°

Substituting the value of A in equation (i) to find the value of B

45Â° + B = 60Â°

B = 60Â° â€“ 45Â°

B = 15Â°

Hence, A = 45Â° and B = 15Â°

### Question 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 Î¸ increases as Î¸ increases.

(iii) The value of cos Î¸ increases as Î¸ increases.

(iv) sin Î¸ = cos Î¸ for all values of Î¸.

(v) cot A is not defined for A = 0Â°.

Solution:

(i) Let us take A = 60Â° and B = 30Â°, then

Substitute the values of A and B in the sin (A + B) formula, we get

sin (A + B) = sin (60Â° + 30Â°) = sin 90Â° = 1 and,

sin A + sin B = sin 60Â° + sin 30Â°

= âˆš3/2 + 1/2 = (âˆš3 + 1 ) / 2, sin(A + B) â‰  sin A + sin B

Since both the values obtained are not equal.

Hence, the statement is false.

(ii) From the values given below, we can see that as angle(theta) increases value also increases.

sin 0Â° = 0, sin 30Â° = 1/2, sin 45Â° = 1/âˆš2, sin 60Â° = âˆš3/2 , sin 90Â° = 1

Thus, the value of sin Î¸ increases as Î¸ increases.

Hence, the statement is true.

(iii) From the values given below, we can see that as angle (theta) increases value decreases.

cos 0Â° = 1, cos 30Â° = âˆš3/2 , cos 45Â° = 1/âˆš2, cos 60Â° = 1/2, cos 90Â° = 0

Thus, the value of cos Î¸ decreases as Î¸ increases.

Hence, the statement given above is false.

(iv) sin Î¸ = cos Î¸, is only true for theta = 45Â°

Therefore, the above statement is false.

(v) As tan 0Â° = 0

cot 0Â° = 1 / tan 0Â°

= 1 / 0 => undefined

Hence, the given statement is true.

## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry: Exercise 8.3

### (i) sin 18Â° / cos 72Â°

Solution:

Since,

cos 72Â°  = cos ( 90Â° – 18Â° ) = sin 18Â°

Therefore,

sin 18Â° / cos 72Â° = sin 18Â° / sin 18Â°  = 1

Hence, sin 18Â° / cos 72Â°  = 1.

### (ii) tan 26Â° / cot 64Â°

Solution:

Since,

cot 64Â°  = cot ( 90Â° – 26Â° ) = tan 26Â°

Therefore,

tan 26Â° / cot 64Â° = tan 26Â° /  tan 26Â°  = 1

Hence, tan 26Â° / cot 64Â°  = 1.

### (iii) cos 48Â° – sin 42Â°

Since,

cos 48Â°  = cos ( 90Â° – 42Â° ) = sin 42Â°

Therefore,

cos 48Â° – sin 42Â° = sin 42Â° –  sin 42Â°  = 0

Hence, cos 48Â° – sin 42Â°  = 0.

### (iv) cosec 31Â° – sec 59Â°

Solution:

Since,

sec 59Â°  = sec ( 90Â° – 31Â° ) = cosec 31Â°

Therefore ,

cosec 31Â° – sec 59Â° = cosec 31Â° – cosec 31Â°  = 0

Hence, cosec 31Â° – sec 59Â°  = 0.

### (i) tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1

Solution:

Let A = tan 48Â° tan 23Â° tan 42Â° tan 67Â°

Since ,

tan 23Â° = tan( 90Â° – 23Â° ) = cot 67Â° and,

tan  42Â° = cot(  90Â° –  42Â° ) = cot  48Â°

Therefore,

A = tan 48Â° cot 67Â° cot  48Â° tan  67Â°

A = 1  (Since, tan BÂ° cot  BÂ° = 1)

Hence, tan 48Â° tan 23Â° tan 42Â° tan 67Â° = 1

### (ii) cos 38Â° cos 52Â° â€“ sin 38Â° sin 52Â° = 0

Let A = cos 38Â° cos 52Â° â€“ sin 38Â° sin 52Â°

Since,

sin 52Â° = sin (90Â° – 38Â°) = cos 38Â° and,

cos  52Â° = cos(90Â° –  52Â°) = sin 38Â°

Therefore,

A = cos 38Â° sin 38Â° â€“ sin 38Â° cos 38Â°

A = 0

Hence, cos 38Â° cos 52Â° â€“ sin 38Â° sin 52Â° = 0.

### Question 3. If tan 2A = cot (A â€“ 18Â°), where 2A is an acute angle, find the value of A.

Solution:

We have,

tan 2A = cot ( A – 18Â° )  —(1)

Since,

tan (2A) = cot ( 90Â° – 2A )  — (2)

Putting (2) in (1),

cot ( 90Â° – 2A ) = cot ( A – 18Â° )

Therefore,

90Â° – 2A = A – 18Â°

3A = 108Â°

A = 36Â°

Hence, A = 36Â°.

### Question 4. If tan A = cot B, prove that A + B = 90Â°.

Solution:

We have,

tan A = cot B —(1)

Since,

tan (A) = cot (90Â° – A)  — (2)

Putting (2) in (1),

cot (90Â° – A) = cot (B)

Therefore,

90Â° – A = B

Hence, A + B = 90Â°.

### Question 5. If sec 4A = cosec (A â€“ 20Â°), where 4A is an acute angle, find the value of A.

Solution:

We have,

sec 4A = cosec ( A – 20Â° )  —(1)

Since,

sec 4A = cosec ( 90Â° – 4A )  — (2)

Putting (2) in (1),

cosec ( 90Â° – 4A ) = cosec ( A  – 20Â° )

Therefore,

90Â° – 4A = A – 20Â°

5A = 110Â°

A = 22Â°

Hence, A = 22Â°.

### Question 6. If A, B,and C are interior angles of a triangle ABC, then show that  sin ((B + C) / 2) = cos (A / 2).

Solution:

Let T = sin ((B + C) / 2) — (1)

A, B and C are the interior angles of triangle ABC, therefore,

A + B + C = 180Â°

Dividing by 2 on both sides

(B + C)/2  = 90Â° – (A / 2) —(2)

Putting (2) on (1)

T = sin (90Â° – (A / 2)

= cos (A / 2)

Hence, sin ((B + C)/2) = cos (A / 2).

### Question 7. Express sin 67Â° + cos 75Â° in terms of trigonometric ratios of angles between 0Â° and 45Â°

Solution:

Let A = sin 67Â° + cos 75Â°

Since,

sin 67Â° = sin(90Â° – 23Â°) = cos (23Â°)

cos 75Â° = cos (90Â° – 15Â°) = sin (15Â°)

Therefore,

sin 67Â° + cos 75Â° = cos 23Â° + sin 15Â°

## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry: Exercise 8.4

### Question 1. Express the trigonometric ratios sin A, sec A, and tan A in terms of cot A

Solution:

(i) sin A

We know that

cosec2A = 1 + cot2A

1/sin2A = 1 + cot2A

sin2A = 1/(1 + cot2A)

sin A = 1/(1+cot2A)1/2

(ii) sec A

sec2A = 1 + tan2A

Sec2A = 1 + 1/cot2A

sec2A = (cot2A + 1) / cot2A

sec A = (cot2A + 1)1/2 / cot A

(iii) tan A

tan A = 1 / cot A

tan A = cot -1 A

### Question 2. Write all the other trigonometric ratios of âˆ A in terms of sec A.

Solution:

(i) cos A

cos A = 1/sec A

(ii) sin A

We know that

sin2A = 1 – cos2A

Also , cos2A = 1 / sec2A

sin2A = 1 – 1 / sec2A

sin2A = (sec2A – 1) / sec2A

sin A = (sec2A – 1)1/2 / sec A

(iii) tan A

We know that

tan2A + 1 = sec2A

tan A = (sec2A – 1)Â½

(iv) cosec A

We know

cosec A = 1/ sinA

cosec A = sec A / (sec2A – 1)Â½

(v) cot A

We know

cot A = cos A / sin A

cot A = (1/sec A) / ((sec2A – 1)1/2 / sec A)

cot A = 1 / (sec2A – 1)1/2

### Question 3. Evaluate:

(i) (sin2 63Â° + sin2 27Â°)/(cos2 17Â° + cos2 73Â°)

(ii) sin 25Â° cos 65Â° + cos 25Â° sin 65Â°

(i) ([sin(90-27)]2 + sin2 27) / ([cos(90-73)]2 + cos2 73)

We know that

sin(90-x) = cos x
cos(90-x) = sin x

(cos2(27) + sin2 27) / (sin2(73) + cos2 73)

Using

sin2A + cos2A = 1

1/1 = 1

(ii) [sin 25 * cos(90-25)] + [cos 25 * sin(90-25)]

Using

sin(90-x) = cos x
cos(90-x) = sin x

= [sin 25 * sin 25] + [cos 25 * cos 25]

= sin2 25 + cos2 25

= 1

### Question 4. Choose the correct option. Justify your choice.

Solution:

(i) 9 sec2 A â€“ 9 tan2 A

(A) 1 (B) 9 (C) 8 (D) 0

Using sec2A – tan2A = 1

9 (sec2A – tan2A ) = 9(1)

Ans (B)

(ii) (1 + tan Î¸ + sec Î¸) (1 + cot Î¸ â€“ cosec Î¸)

(A) 0 (B) 1 (C) 2 (D) â€“1

Simplifying all ratios

= (1 + sinÎ¸/cosÎ¸ + 1/cosÎ¸) (1 + cosÎ¸/sinÎ¸ – 1/sinÎ¸)

= ((cosÎ¸ + sinÎ¸ + 1)/ cosÎ¸) ((sinÎ¸ + cosÎ¸ – 1 )/sinÎ¸)

= ((cosÎ¸ + sinÎ¸)2 – 1) / (sinÎ¸ cosÎ¸)

= (1 + 2*cosÎ¸*sinÎ¸ – 1) / (sinÎ¸ cosÎ¸)

= 2

Ans (C)

(iii) (sec A + tan A) * (1 â€“ sin A)

(A) sec A (B) sin A (C) cosec A (D) cos A

Simplifying sec A and tan A

= (1/cos A + sin A/cos A)*(1 – sin A)

= ((1 + sin A)/cos A)*(1 – sin A)

= (1 – sin2A)/cos A

= cos2A / cos A

= cos A

Ans (D)

(iv) (1 + tan2A) / (1 + cot2A)

(A) sec2A (B) â€“1 (C) cot2A (D) tan2A

Simplifying tan A and cot A

= (1 + (sin2A / cos2A)) / (1 + (cos2A / sin2A))

= ((cos2A + sin2A) / cos2A) / ((cos2A + sin2A) / sin2A)

= sin2A / cos2A

= tan2A

Ans (D)

### Question 5. Prove the following identities, where the angles involved are acute angles for which the expressions are defined.

Solution:

(i) (cosec Î¸ â€“ cot Î¸)2 = (1 – cosÎ¸) / (1 + cosÎ¸)

Solving LHS

Simplifying cosec Î¸ and cot Î¸

= (1-cos Î¸)2 / sin2Î¸

= (1-cos Î¸)2 / (1-cos2Î¸)

Using a2 – b2 = (a+b)*(a-b)

= (1-cos Î¸)2 / [(1-cos Î¸)*(1+cos Î¸)]

= (1-cos Î¸) / (1+cos Î¸) = RHS

Hence Proved

(ii) (cos A / (1+sin A) + ((1+sin A) / cos A) = 2 sec A

Solving LHS

Taking LCM

= (cos2A + (1+sin A)2) / ((1+sin A) cos A)

= (cos2A + 1 + sin2A + 2 sin A ) / ((1 + sin A)*cos A)

Using sin2A + cos2A = 1

= (2 + 2*sin A) / ((1+sin A)*cos A)

= (2*(1 + sin A)) / ((1 + sin A)*cos A)

= 2 / cos A

= 2 sec A = RHS

Hence Proved

(iii) (tan Î¸ / (1 – cot Î¸)) + (cot Î¸ / (1 – tan Î¸)) = 1 + sec Î¸*cosec Î¸

Solving LHS

Changing tan Î¸ and cot Î¸ in terms of sin Î¸ and cos Î¸ and simplifying

= ((sin2Î¸) / (cos Î¸ *(sin Î¸-cos Î¸))) + ((cos2Î¸ ) / (sin Î¸ *(sin Î¸-cos Î¸)))

= (1 / (sin Î¸-cos Î¸)) * [(sin3Î¸ – cos3Î¸) / (sin Î¸ * cos Î¸)]

= (1 / (sin Î¸ – cos Î¸)) * [ ((sin Î¸ – cos Î¸) * ( sin2Î¸ + cos2Î¸ + sin Î¸ * cos Î¸ ))/(sin Î¸ *cos Î¸)]

= (1+sin Î¸*cos Î¸) / (sin Î¸*cos Î¸)

= sec Î¸*cosec Î¸ + 1 = RHS

Hence Proved

(iv) (1 + sec A) / sec A = sin2A / (1 – cos A)

Solving LHS

= cos A + 1

Solving RHS

= (1 – cos2A) / (1 – cos A)

= (1 – cos A) * (1 + cos A) / (1 – cos A)

= 1 + cos A = RHS

Hence Proved

(v) (cos A – sin A + 1) / (cos A + sin A – 1) = cosec A + cot A using the identity cosec2A = 1 + cot2A

Solving LHS

Multiplying numerator and denominator by (cot A – 1 + cosec A)

= (cot2A + 1 + cosec2A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot2A – (1 + cosec2A – 2*cosec A))

= (2*cosec2A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot2A – 1 – cosec2A + 2*cosec A)

= (2* cosec A *(cosec A + cot A) – 2*(cosec A + cot A)) / (cot2A – 1 – cosec2A + 2*cosec A)

= ((cosec A + cot A) * (2*cosec A – 2 )) / (2*cosec A – 2)

= cosec A + cot A = RHS

Hence Proved

(vi) [(1 + sin A) / (1 – sin A)]Â½ = sec A + tan A

Solving LHS

Multiplying numerator and denominator by (1+sinA)

= [((1 + sin A)*(1 + sin A)) / ((1 – sin A)*(1 + sin A))]Â½

= (1 + sin A) / (1 – sin2A)Â½

= (1 + sin A) / (cos2A)1/2

= (1 + sin A) / (cos A)

= sec A + tan A = RHS

Hence Proved

(vii) (sin Î¸ – 2 sin3Î¸) / (2 cos3Î¸ – cos Î¸) = tan Î¸

Solving LHS

= (sin Î¸ * (1 – 2*sin2Î¸)) / (cos Î¸ * (2*cos2Î¸ – 1))

= (sin Î¸ * (1 – 2*sin2Î¸ )) / (cos Î¸ * (2*(1 – sin2Î¸) – 1))

= (sin Î¸ *(1 – 2*sin2Î¸)) / (cos Î¸ * (1 – 2*sin2Î¸))

= tan Î¸ = RHS

Hence Proved

(viii) (sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2A + cot2A

Solving LHS

= sin2A + cosec2A + 2*sin A *cosec A + cos2A + sec2A + 2*cos A *sec A

We know that cosec A = 1 / sin A

= 1 + 1 + cot2A + 1 + tan2A + 2 + 2

= 7 + tan2A + cot2A = RHS

Hence Proved

(ix) (cosec A â€“ sin A)*(sec A â€“ cos A) = 1 / (tan A + cot A)

Solving LHS

= ((1/sin A) – sin A) * ((1/cos A) – cos A)

= ((1 – sin2A) / sin A) * ((1 – cos2A) / cos A)

= (cos2A * sin2A) / (sin A * cos A)

= sin A * cos A

Solving RHS

Simplifying tan A and cot A

= (sin A * cos A) / ( sin2A + cos2A)

= sin A * cos A = RHS

Hence Proved

(x) (1 + tan2A) / (1 + cot2A ) = [(1 – tan A) / (1 – cot A)]2 = tan2A

Solving LHS

Changing cot A = 1 / tan A

= (tan2A * (1 + tan2A)) / (1 + tan2A) = tan2A = RHS

= [(1 – tan A) / (1 – cot A)]2 = (1 + tan2A – 2*tan A) / (1 + cot2A – 2*cot A)

= (sec2A – 2*tan A) / (cosec2A – 2*cot A)

Solving this we get

= tan2A

Hence Proved

## Key Features of NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry:

• NCERT Solutions are created for each chapter including Class 10 Maths Chapter 8 Introduction to Trigonometry.
• These solutions provide in-depth solutions to the problems encountered by students in their NCERT Class 10 Maths textbook.
• These solutions are very accurate and comprehensive, which can help students of Class 10 prepare for any academic as well as competitive exams.

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## NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry- FAQs

### Q1: Why is it important to learn Class 10 Maths Chapter 8 Introduction to Trigonometry?

Trigonometry is used to set directions. It also tells you the direction to point the compass in order to travel straight forward. To find a certain location, it is used in navigation. It is also used to calculate the separation between a location in the water and the coast.

### Q2: What topics are covered in NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry?

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry covers topics such as trigonometric ratios and their calculation with the help of Pythagoras theorem, calculation of specific angles like 0 degrees, 30 degrees, 45 degrees, and 90 degrees and the use of the trigonometric table, complementary angle, and trigonometric identities.

### Q3: How can NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry help me?

NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry can help you solve the NCERT exercise without any limitations. If you are stuck on a problem you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.

### Q4: How many exercises are there in NCERT Class 10 Maths Chapter 8 Introduction to Trigonometry?

There are 4 exercises in the Class 10 NCERT Maths Chapter 8 Introduction to Trigonometry which covers all the important topics and sub-topics.

### Q5: Where can I find NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry?

You can find NCERT Solutions for Class 10 Maths Chapter 8 Introduction to Trigonometry in this article, created by our team of experts at GeeksforGeeks.

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