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:
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
Question 1. In ∆ ABC, right-angled at B, AB = 24 cm, BC = 7 cm. Determine:
(i) sin A, cos A (ii) sin C, cos C
Solution:
Using Pythagoras theorem for ΔABC
AC2 = AB2 + BC2
= (24 cm)2 + (7 cm)2
= (576 + 49) cm2
= 625 cm2
∴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 – cot R.
Solution:
Applying Pythagoras theorem for ΔPQR, 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 = opp/adj
tan P = 5/12
cot R = adj/opp
cot R = 5/12
tan P – cot R = 5/12 – 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 – (3/4)2 = 1 – 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)2 = sec2A
289/64 = sec2A
sec A = 17/8
We know, cos2A = 1/sec2A
cos2A = 64/289
sin2A = 1 – cos2A
sin2A = 225/289
sin A = 15/17
Question 5: Given sec θ = 13/12, calculate all other trigonometric ratios.
Solution.
Using Pythagoras theorem,
sin θ = 5/13
cos θ = 12/13
tan θ = 5/12
cosec θ = 13/5
cot θ = 12/5
Question 6: If ∠A and ∠B are acute angles such that cos A = cos B, then show that ∠A = ∠B.
Solution:
Let us consider a ΔABC in which CD ⊥ AB.
It is given that cos A = cos B
AD/AC = BD/BC … (1)
We need to prove ∠A = ∠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),
CD||BP
∠ACD = ∠CPB (Corresponding angles) … (3)
And, ∠BCD = ∠CBP (Alternate interior angles) … (4)
By construction, we have BC = CP.
∴ ∠CBP = ∠CPB (Angle opposite to equal sides of a triangle) … (5)
From equations (3), (4), and (5), we obtain
∠ACD = ∠BCD … (6)
In ΔCAD and ΔCBD,
∠ACD = ∠BCD (Using equation (6))
∠CDA = ∠CDB (Both 90°)
Therefore, the remaining angles should be equal.
∴∠CAD = ∠CBD
⇒ ∠A = ∠B
Question 7: If cot θ = 7/8, evaluate
(i) ((1 + sinθ) * (1 – sinθ))/(1 + cosθ) * (1 – cosθ)))
(ii) cot2θ
Solution:
(i) Using (a + b) * (a – b) = a2 – b2 in numerator and denominator
We get
(1 – sin2θ)/(1 – cos2θ)
Using sin2θ + cos2θ = 1
We get
cos2θ/sin2θ = cot2θ
Now
cot2θ = (7/8)2 = 49/64
(ii) cot2θ = (7/8)2 = 49/64
Question 8. If 3 cot A = 4, Check whether (1 – tan2A)/(1 + tan2A) = cos2A – sin2A
Solution.
We know that, tanA = sinA / cosA ….(1)
Using (1) on L.H.S
= (1 – sin2A/cos2A)/(1 + sin2A/cos2A)
which on rearranging becomes
= (cos2A – sin2A)/(cos2A + sin2A)
Using the identity,
cos2A + sin2A = 1
LHS becomes
= (cos2A – sin2A)
This is equal to RHS.
LHS = RHS (for every value of cot A)
Hence, Proved.
Question 9: In ΔABC, right-angled at B. If tan A = 1/(31/2), find the value of
(i) sin A cos C + cos A sin C
(ii) cos A cos C − 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 − sin A sin C
Substituting the values
= (31/2/2) * (1/2) – (1/2) * (31/2/2)
= 31/2/4 – 31/2/4
= 0
Question 10: In ΔPQR, 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 − x cm
Applying Pythagoras theorem in ΔPQR, we obtain
PR2 = PQ2 + QR2
x2 = (5)2 + (25 − x)2
x2 = 25 + 625 + x2 − 50x
50x = 650
x = 13
Therefore, PR = 13 cm
QR = (25 − 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 θ = 4/3, for some angle θ
Solution:
(i) Consider a ΔPQR, 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 ΔABC with AB = 5 cm, AC = 12 cm and BC = x cm
Using Pythagoras theorem in ΔABC
(AB)2 + (BC)2 = (AC)2
52 + x2 = 122
x = (144 – 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 ∠A.
Hence, the given statement is false.
(v) sin θ = 4/3
In a right-angled triangle, hypotenuse is always greater than the remaining two sides. Therefore, such value of sin θ 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 60° cos 30° + sin 30° cos 60°
Solution:
Formulas to be used : sin 30° = 1/2
cos 30° = √3/2
sin 60° = 3/2
cos 60° = 1/2
=> (√3/2) * (√3/2) + (1/2) * (1/2)
=> 3/4 +1/4
=> 4 /4
=> 1
(ii) 2 tan245° + cos230° – sin260°
Solution:
Formulas to be used : sin 60° = √3/2
cos 30° = √3/2
tan 45° = 1
=> 2(1)(1) + (√3/2)(√3/2)-(√3/2)(√3/2)
=> 2 + 3/4 – 3/4
=> 2
(iii) cos 45°/(sec 30°+cosec 30°)
Solution:
Formulas to be used : cos 45° = 1/√2
sec 30° = 2/√3
cosec 30° = 2
=> 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 45º – cosec 60°)/(sec 30° + cos 60° + cot 45°)
Solution:
Formulas to be used : sin 30° = 1/2
tan 45° = 1
cosec 60° = 2/√3
sec 30° = 2/√3
cos 60° = 1/2
cot 45° = 1
=> (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)
=> (27+16-24√3) / (27-16)
=> (43-24√3)/11
(v) (5cos260° + 4sec2 30° – tan245°)/(sin230° + cos²30°)
Solution:
Formulas to be used : cos 60° = 1/2
sec 30° = 2/√3
tan 45° = 1
sin 30° = 1/2
cos 30° = √3/2
=> 5(1/2)2+4(2/√3)²-1²/(1/2)+(√3/2)
=> (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 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°
Solution:
(i) In the given equation, substituting the value of tan 30°
As tan 30° = 1/√3
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°
The ans is sin 60°.
The correct option is (A).
(ii) In the given equation, substituting the of tan 45°
As tan 45° = 1
1-tan245°/1+tan245° = (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 = 0°
sin 2A = sin 0° = 0
2 sin A = 2 sin 0° = 2 × 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 = 0°
The correct option is (A).
(iv) As tan 30° = 1/√3
2tan30°/1-tan230° = 2(1/√3)/1-(1/√3)2
=> (2/√3)/(1-1/3) = (2/√3)/(2/3) = √3 = tan 60°
The correct option is (C).
Question 3. If tan (A + B) = √3 and tan (A – B) = 1/√3, 0° < A + B ≤ 90°; A > B, find A and B.
Solution:
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)
Now add the equation (i) and (ii), we get
A + B + A – B = 60° + 30°
A= 45°
Substituting the value of A in equation (i) to find the value of B
45° + B = 60°
B = 60° – 45°
B = 15°
Hence, A = 45° and B = 15°
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 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°.
Solution:
(i) Let us take A = 60° and B = 30°, then
Substitute the values of A and B in the sin (A + B) formula, we get
sin (A + B) = sin (60° + 30°) = sin 90° = 1 and,
sin A + sin B = sin 60° + sin 30°
= √3/2 + 1/2 = (√3 + 1 ) / 2, sin(A + B) ≠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 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.
Hence, the statement is true.
(iii) From the values given below, we can see that as angle (theta) increases value decreases.
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.
Hence, the statement given above is false.
(iv) sin θ = cos θ, is only true for theta = 45°
Therefore, the above statement is false.
(v) As tan 0° = 0
cot 0° = 1 / tan 0°
= 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 18° / cos 72°
Solution:
Since,
cos 72° = cos ( 90° – 18° ) = sin 18°
Therefore,
sin 18° / cos 72° = sin 18° / sin 18° = 1
Hence, sin 18° / cos 72° = 1.
(ii) tan 26° / cot 64°
Solution:
Since,
cot 64° = cot ( 90° – 26° ) = tan 26°
Therefore,
tan 26° / cot 64° = tan 26° / tan 26° = 1
Hence, tan 26° / cot 64° = 1.
(iii) cos 48° – sin 42°
Since,
cos 48° = cos ( 90° – 42° ) = sin 42°
Therefore,
cos 48° – sin 42° = sin 42° – sin 42° = 0
Hence, cos 48° – sin 42° = 0.
(iv) cosec 31° – sec 59°
Solution:
Since,
sec 59° = sec ( 90° – 31° ) = cosec 31°
Therefore ,
cosec 31° – sec 59° = cosec 31° – cosec 31° = 0
Hence, cosec 31° – sec 59° = 0.
Question 2. Show that:
(i) tan 48° tan 23° tan 42° tan 67° = 1
Solution:
Let A = tan 48° tan 23° tan 42° tan 67°
Since ,
tan 23° = tan( 90° – 23° ) = cot 67° and,
tan 42° = cot( 90° – 42° ) = cot 48°
Therefore,
A = tan 48° cot 67° cot 48° tan 67°
A = 1 (Since, tan B° cot B° = 1)
Hence, tan 48° tan 23° tan 42° tan 67° = 1
(ii) cos 38° cos 52° – sin 38° sin 52° = 0
Let A = cos 38° cos 52° – sin 38° sin 52°
Since,
sin 52° = sin (90° – 38°) = cos 38° and,
cos 52° = cos(90° – 52°) = sin 38°
Therefore,
A = cos 38° sin 38° – sin 38° cos 38°
A = 0
Hence, cos 38° cos 52° – sin 38° sin 52° = 0.
Question 3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.
Solution:
We have,
tan 2A = cot ( A – 18° ) —(1)
Since,
tan (2A) = cot ( 90° – 2A ) — (2)
Putting (2) in (1),
cot ( 90° – 2A ) = cot ( A – 18° )
Therefore,
90° – 2A = A – 18°
3A = 108°
A = 36°
Hence, A = 36°.
Question 4. If tan A = cot B, prove that A + B = 90°.
Solution:
We have,
tan A = cot B —(1)
Since,
tan (A) = cot (90° – A) — (2)
Putting (2) in (1),
cot (90° – A) = cot (B)
Therefore,
90° – A = B
Hence, A + B = 90°.
Question 5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.
Solution:
We have,
sec 4A = cosec ( A – 20° ) —(1)
Since,
sec 4A = cosec ( 90° – 4A ) — (2)
Putting (2) in (1),
cosec ( 90° – 4A ) = cosec ( A – 20° )
Therefore,
90° – 4A = A – 20°
5A = 110°
A = 22°
Hence, A = 22°.
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 = 180°
Dividing by 2 on both sides
(B + C)/2 = 90° – (A / 2) —(2)
Putting (2) on (1)
T = sin (90° – (A / 2)
= cos (A / 2)
Hence, sin ((B + C)/2) = cos (A / 2).
Question 7. Express sin 67° + cos 75° in terms of trigonometric ratios of angles between 0° and 45°
Solution:
Let A = sin 67° + cos 75°
Since,
sin 67° = sin(90° – 23°) = cos (23°)
cos 75° = cos (90° – 15°) = sin (15°)
Therefore,
sin 67° + cos 75° = cos 23° + sin 15°
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 ∠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)½
(iv) cosec A
We know
cosec A = 1/ sinA
cosec A = sec A / (sec2A – 1)½
(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 63° + sin2 27°)/(cos2 17° + cos2 73°)
(ii) sin 25° cos 65° + cos 25° sin 65°
(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 – 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 θ + sec θ) (1 + cot θ – cosec θ)
(A) 0 (B) 1 (C) 2 (D) –1
Simplifying all ratios
= (1 + sinθ/cosθ + 1/cosθ) (1 + cosθ/sinθ – 1/sinθ)
= ((cosθ + sinθ + 1)/ cosθ) ((sinθ + cosθ – 1 )/sinθ)
= ((cosθ + sinθ)2 – 1) / (sinθ cosθ)
= (1 + 2*cosθ*sinθ – 1) / (sinθ cosθ)
= 2
Ans (C)
(iii) (sec A + tan A) * (1 – 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) –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 θ – cot θ)2 = (1 – cosθ) / (1 + cosθ)
Solving LHS
Simplifying cosec θ and cot θ
= (1-cos θ)2 / sin2θ
= (1-cos θ)2 / (1-cos2θ)
Using a2 – b2 = (a+b)*(a-b)
= (1-cos θ)2 / [(1-cos θ)*(1+cos θ)]
= (1-cos θ) / (1+cos θ) = 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 θ / (1 – cot θ)) + (cot θ / (1 – tan θ)) = 1 + sec θ*cosec θ
Solving LHS
Changing tan θ and cot θ in terms of sin θ and cos θ and simplifying
= ((sin2θ) / (cos θ *(sin θ-cos θ))) + ((cos2θ ) / (sin θ *(sin θ-cos θ)))
= (1 / (sin θ-cos θ)) * [(sin3θ – cos3θ) / (sin θ * cos θ)]
= (1 / (sin θ – cos θ)) * [ ((sin θ – cos θ) * ( sin2θ + cos2θ + sin θ * cos θ ))/(sin θ *cos θ)]
= (1+sin θ*cos θ) / (sin θ*cos θ)
= sec θ*cosec θ + 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)]½ = 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))]½
= (1 + sin A) / (1 – sin2A)½
= (1 + sin A) / (cos2A)1/2
= (1 + sin A) / (cos A)
= sec A + tan A = RHS
Hence Proved
(vii) (sin θ – 2 sin3θ) / (2 cos3θ – cos θ) = tan θ
Solving LHS
= (sin θ * (1 – 2*sin2θ)) / (cos θ * (2*cos2θ – 1))
= (sin θ * (1 – 2*sin2θ )) / (cos θ * (2*(1 – sin2θ) – 1))
= (sin θ *(1 – 2*sin2θ)) / (cos θ * (1 – 2*sin2θ))
= tan θ = 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 – sin A)*(sec A – 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.
Also Check:
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.
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...