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 stepbystep 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 202324 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
Question 1. In âˆ† ABC, rightangled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A (ii) sin C, cos C
Solution:
Using Pythagoras theorem for Î”ABC
AC^{2} = AB^{2} + BC^{2}
= (24 cm)^{2} + (7 cm)^{2}
= (576 + 49) cm^{2}
= 625 cm^{2}
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 = adj/hyp
cos C = 7/25
Question 2. In Fig., find tan P â€“ cot R.
Solution:
Applying Pythagoras theorem for Î”PQR, we obtain
PR^{2} = PQ^{2} + QR^{2}
(13 cm)^{2} = (12 cm)^{2} + QR^{2}
169 cm^{2} = 144 cm^{2} + QR^{2}
25 cm^{2} = QR^{2}
QR = 5 cm
tan P = opp/adj
tan P = 5/12
cot R = adj/opp
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 sin^{2}A + cos^{2}A = 1
(3/4)^{2} + cos^{2}A = 1
cos^{2}A = 1 â€“ (3/4)^{2} = 1 â€“ 9/16
cos A = 7^{1/2}/4
tan A = sin A/cos A
tan A = (3/4)/(7^{1/2}/4)
tan A = 3/7^{1/2}
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 + tan^{2}A = sec^{2}A
1 + (15/8)^{2 }= sec^{2}A
289/64 = sec^{2}A
sec A = 17/8
We know, cos^{2}A = 1/sec^{2}A
cos^{2}A = 64/289
sin^{2}A = 1 â€“ cos2A
sin^{2}A = 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
AD/AC = BD/BC â€¦ (1)
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/BC
AD/BD = AC/CP (BC = CP by construction)
By using the converse of B.P.T (Basic Proportionality Theorem),
CDBP
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)
In Î”CAD and Î”CBD,
âˆ ACD = âˆ BCD (Using equation (6))
âˆ CDA = âˆ CDB (Both 90Â°)
Therefore, the remaining angles should be equal.
âˆ´âˆ CAD = âˆ CBD
â‡’ âˆ A = âˆ B
Question 7: If cot Î¸ = 7/8, evaluate
(i) ((1 + sinÎ¸) * (1 â€“ sinÎ¸))/(1 + cosÎ¸) * (1 â€“ cosÎ¸)))
(ii) cot^{2}Î¸
Solution:
(i) Using (a + b) * (a â€“ b) = a^{2} â€“ b^{2} in numerator and denominator
We get
(1 â€“ sin^{2}Î¸)/(1 â€“ cos^{2}Î¸)
Using sin^{2}Î¸ + cos^{2}Î¸ = 1
We get
cos^{2}Î¸/sin^{2}Î¸ = cot^{2}Î¸
Now
cot^{2}Î¸ = (7/8)2 = 49/64
(ii) cot^{2}Î¸ = (7/8)2 = 49/64
Question 8. If 3 cot A = 4, Check whether (1 â€“ tan^{2}A)/(1 + tan^{2}A) = cos^{2}A â€“ sin^{2}A
Solution.
We know that, tanA = sinA / cosA â€¦.(1)
Using (1) on L.H.S
= (1 â€“ sin^{2}A/cos^{2}A)/(1 + sin^{2}A/cos^{2}A)
which on rearranging becomes
= (cos^{2}A â€“ sin^{2}A)/(cos^{2}A + sin^{2}A)
Using the identity,
cos^{2}A + sin^{2}A = 1
LHS becomes
= (cos^{2}A â€“ sin^{2}A)
This is equal to RHS.
LHS = RHS (for every value of cot A)
Hence, Proved.
Question 9: In Î”ABC, rightangled at B. If tan A = 1/(3^{1/2}), find the value of
(i) sin A cos C + cos A sin C
(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 = 3^{1/2}/2
cos A = 3^{1/2}/2
cos C = 1/2
Now, (i) sin A cos C + cos A sin C
Substituting the values
= (1/2) * (1/2) + (3^{1/2}/2) * (3^{1/2}/2)
= 1/4 + 3/4
= 1
Now, (ii) cos A cos C âˆ’ sin A sin C
Substituting the values
= (3^{1/2}/2) * (1/2) â€“ (1/2) * (3^{1/2}/2)
= 3^{1/2}/4 â€“ 3^{1/2}/4
= 0
Question 10: In Î”PQR, rightangled 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
PR^{2} = PQ^{2} + QR^{2}
x^{2} = (5)^{2} + (25 âˆ’ x)^{2}
x^{2} = 25 + 625 + x^{2} 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
Question 11. State whether the following are true or false. Justify your answer.
(i) The value of tan P 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 Î¸ = 4/3, for some angle Î¸
Solution:
(i) Consider a Î”PQR, rightangled 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 rightangled 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
Question 1. Evaluate the following :
(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 tan^{2}45Â° + cos^{2}30Â° â€“ sin^{2}60Â°
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(2âˆš62âˆš2)/(2âˆš6+2âˆš2)(2âˆš62âˆš2)
=> 2âˆš3(âˆš6âˆš2) / (2âˆš6)Â²(2âˆš2)Â²
=> 2âˆš3(âˆš6âˆš2)/(248) = 2 âˆš3(âˆš6âˆš2)/16
=> âˆš3(âˆš6âˆš2)/8
=> (âˆš18âˆš6)/8
=> (3âˆš2âˆš6)/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+12/âˆš3) / (2/âˆš3+1/2+1)
=> (3/22/âˆš3)/(3/2+2/âˆš3)
=> (3âˆš34/2 âˆš3)/(3âˆš3+4/2 âˆš3)
=> (3âˆš34)(3âˆš34)/(3âˆš3+4)(3âˆš34)
=> (27+1624âˆš3) / (2716)
=> (4324âˆš3)/11
(v) (5cos^{2}60Â° + 4sec^{2} 30Â° – tan^{2}45Â°)/(sin^{2}30Â° + 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)Â²1Â²/(1/2)+(âˆš3/2)
=> (5/4+16/31) / (1/4+3/4)
=> (15+6412) / 12/(4/4)
=> 67/12
Question 2. Choose the correct option and justify your choice:
(i) 2tan 30Â°/1+tan^{2}30Â° =
(A) sin 60Â° (B) cos 60Â° (C) tan 60Â° (D) sin 30Â°
(ii) 1tan^{2}45Â°/1+tan^{2}45Â° =
(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°/1tan^{2}30Â° =
(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+tan^{2}30Â° = 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
1tan^{2}45Â°/1+tan^{2}45Â° = (11^{2})/(1+1^{2})
= 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°/1tan^{2}30Â° = 2(1/âˆš3)/1(1/âˆš3)^{2}
=> (2/âˆš3)/(11/3) = (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
Question 1. Evaluate:
(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.
Question 2. Show that:
(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
cosec^{2}A = 1 + cot^{2}A
1/sin^{2}A = 1 + cot^{2}A
sin^{2}A = 1/(1 + cot^{2}A)
sin A = 1/(1+cot^{2}A)^{1/2}
(ii) sec A
sec^{2}A = 1 + tan^{2}A
Sec^{2}A = 1 + 1/cot^{2}A
sec^{2}A = (cot^{2}A + 1) / cot^{2}A
sec A = (cot^{2}A + 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
sin^{2}A = 1 – cos^{2}A
Also , cos^{2}A = 1 / sec^{2}A
sin^{2}A = 1 – 1 / sec^{2}A
sin^{2}A = (sec^{2}A – 1) / sec^{2}A
sin A = (sec^{2}A – 1)^{1/2} / sec A
(iii) tan A
We know that
tan^{2}A + 1 = sec^{2}A
tan A = (sec^{2}A – 1)Â½
(iv) cosec A
We know
cosec A = 1/ sinA
cosec A = sec A / (sec^{2}A – 1)Â½
(v) cot A
We know
cot A = cos A / sin A
cot A = (1/sec A) / ((sec^{2}A – 1)^{1/2} / sec A)
cot A = 1 / (sec^{2}A – 1)^{1/2}
Question 3. Evaluate:
(i) (sin^{2} 63Â° + sin^{2} 27Â°)/(cos^{2} 17Â° + cos^{2} 73Â°)
(ii) sin 25Â° cos 65Â° + cos 25Â° sin 65Â°
(i) ([sin(9027)]^{2} + sin^{2} 27) / ([cos(9073)]^{2} + cos^{2} 73)
We know that
sin(90x) = cos x
cos(90x) = sin x(cos^{2}(27) + sin^{2} 27) / (sin^{2}(73) + cos^{2} 73)
Using
sin^{2}A + cos^{2}A = 1
1/1 = 1
(ii) [sin 25 * cos(9025)] + [cos 25 * sin(9025)]
Using
sin(90x) = cos x
cos(90x) = sin x= [sin 25 * sin 25] + [cos 25 * cos 25]
= sin^{2} 25 + cos^{2} 25
= 1
Question 4. Choose the correct option. Justify your choice.
Solution:
(i) 9 sec^{2 }A â€“ 9 tan^{2} A
(A) 1 (B) 9 (C) 8 (D) 0
Using sec^{2}A – tan^{2}A = 1
9 (sec^{2}A – tan^{2}A ) = 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 – sin^{2}A)/cos A
= cos^{2}A / cos A
= cos A
Ans (D)
(iv) (1 + tan^{2}A) / (1 + cot^{2}A)
(A) sec^{2}A (B) â€“1 (C) cot^{2}A (D) tan^{2}A
Simplifying tan A and cot A
= (1 + (sin^{2}A / cos^{2}A)) / (1 + (cos^{2}A / sin^{2}A))
= ((cos^{2}A + sin^{2}A) / cos^{2}A) / ((cos^{2}A + sin^{2}A) / sin^{2}A)
= sin^{2}A / cos^{2}A
= tan^{2}A
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¸
= (1cos Î¸)^{2} / sin^{2}Î¸
= (1cos Î¸)^{2} / (1cos^{2}Î¸)
Using a^{2} – b^{2} = (a+b)*(ab)
= (1cos Î¸)^{2} / [(1cos Î¸)*(1+cos Î¸)]
= (1cos 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
= (cos^{2}A + (1+sin A)^{2}) / ((1+sin A) cos A)
= (cos^{2}A + 1 + sin^{2}A + 2 sin A ) / ((1 + sin A)*cos A)
Using sin^{2}A + cos^{2}A = 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
= ((sin^{2}Î¸) / (cos Î¸ *(sin Î¸cos Î¸))) + ((cos^{2}Î¸ ) / (sin Î¸ *(sin Î¸cos Î¸)))
= (1 / (sin Î¸cos Î¸)) * [(sin^{3}Î¸ – cos^{3}Î¸) / (sin Î¸ * cos Î¸)]
= (1 / (sin Î¸ – cos Î¸)) * [ ((sin Î¸ – cos Î¸) * ( sin^{2}Î¸ + cos^{2}Î¸ + 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 = sin^{2}A / (1 – cos A)
Solving LHS
= cos A + 1
Solving RHS
= (1 – cos^{2}A) / (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 cosec^{2}A = 1 + cot^{2}A
Solving LHS
Multiplying numerator and denominator by (cot A – 1 + cosec A)
= (cot^{2}A + 1 + cosec^{2}A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot^{2}A – (1 + cosec^{2}A – 2*cosec A))
= (2*cosec^{2}A – 2*cot A – 2*cosec A + 2*cot A*cosec A) / (cot^{2}A – 1 – cosec^{2}A + 2*cosec A)
= (2* cosec A *(cosec A + cot A) – 2*(cosec A + cot A)) / (cot^{2}A – 1 – cosec^{2}A + 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 – sin^{2}A)^{Â½}
= (1 + sin A) / (cos^{2}A)^{1/2}
= (1 + sin A) / (cos A)
= sec A + tan A = RHS
Hence Proved
(vii) (sin Î¸ – 2 sin^{3}Î¸) / (2 cos^{3}Î¸ – cos Î¸) = tan Î¸
Solving LHS
= (sin Î¸ * (1 – 2*sin^{2}Î¸)) / (cos Î¸ * (2*cos^{2}Î¸ – 1))
= (sin Î¸ * (1 – 2*sin^{2}Î¸ )) / (cos Î¸ * (2*(1 – sin^{2}Î¸) – 1))
= (sin Î¸ *(1 – 2*sin^{2}Î¸)) / (cos Î¸ * (1 – 2*sin^{2}Î¸))
= tan Î¸ = RHS
Hence Proved
(viii) (sin A + cosec A)^{2 }+ (cos A + sec A)^{2} = 7 + tan^{2}A + cot^{2}A
Solving LHS
= sin^{2}A + cosec^{2}A + 2*sin A *cosec A + cos^{2}A + sec^{2}A + 2*cos A *sec A
We know that cosec A = 1 / sin A
= 1 + 1 + cot^{2}A + 1 + tan^{2}A + 2 + 2
= 7 + tan^{2}A + cot^{2}A = 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 – sin^{2}A) / sin A) * ((1 – cos^{2}A) / cos A)
= (cos^{2}A * sin^{2}A) / (sin A * cos A)
= sin A * cos A
Solving RHS
Simplifying tan A and cot A
= (sin A * cos A) / ( sin^{2}A + cos^{2}A)
= sin A * cos A = RHS
Hence Proved
(x) (1 + tan^{2}A) / (1 + cot^{2}A ) = [(1 – tan A) / (1 – cot A)]^{2} = tan^{2}A
Solving LHS
Changing cot A = 1 / tan A
= (tan^{2}A * (1 + tan^{2}A)) / (1 + tan^{2}A) = tan^{2}A = RHS
= [(1 – tan A) / (1 – cot A)]^{2} = (1 + tan^{2}A – 2*tan A) / (1 + cot^{2}A – 2*cot A)
= (sec^{2}A – 2*tan A) / (cosec^{2}A – 2*cot A)
Solving this we get
= tan^{2}A
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 indepth 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.
Also Check:
 NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers
 NCERT Solutions for Class 10 Maths Chapter 2 Polynomials
 NCERT Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
 NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equation
 NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions
 NCERT Solutions for Class 10 Maths Chapter 6 Triangles
 NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
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 subtopics.
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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