Careers360 Logo
NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

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

Updated on Apr 30, 2025 03:22 PM IST | #CBSE Class 10th

Trigonometric identities provide a foundation that enables quick solutions to multiple mathematical expressions. This exercise demonstrates applying basic trigonometric ratios of sine, cosine and tangent when performing standard identity simplifications and verifications. The linkage among different trigonometric functions becomes possible through these identities, which enable us to show complex expressions in separate steps. Learning trigonometric connections helps students develop their logical reasoning skills, leading to advanced trigonometric abilities needed for engineering, architecture and physical science fields.

This Story also Contains
  1. NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.3
  2. Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.3
  3. Topics covered in Chapter 8 Introduction to Trigonometry: Exercise 8.3
  4. NCERT Solutions of Class 10 Subject Wise
  5. NCERT Exemplar Solutions of Class 10 Subject-Wise
NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry
NCERT Solutions for Exercise 8.4 Class 10 Maths Chapter 8 - Introduction to Trigonometry

Exercise 8.3 in the NCERT Solutions for Class 10 presents three important identities which state sin2θ+cos2θ=1 alongside 1+tan2θ=sec2θ and 1+cot2θ=csc2θ. The relationships teach students methods to simplify problems through systematic verification of given expressions. The exercise functions as an essential tool for solidifying the knowledge described in NCERT Books. This exercise represents a fundamental requirement for solving trigonometric problems, which progress into height and distance calculations and advanced mathematical concepts.

NCERT Solutions Class 10 Maths Chapter 8: Exercise 8.3

Download PDF


Access Solution of Introduction to Trigonometry Class 10 Chapter 8 Exercise: 8.3

Q 1: Express the trigonometric ratios sinA,secA and tanA in terms of cotA .

Answer:

We know that csc2Acot2A=1
(i)
1sin2A=1+cot2Asin2A=11+cot2AsinA=11+cot2A

(ii) We know the identity of
sec2Atan2A=11cos2A=1+tan2A=1+1cot2Acot2A1+cot2A=cos2AcotA1+cot=cosA

(iii) tanA=1cotA

Q 2: Write all the other trigonometric ratios of A in terms of secA .

Answer:

We know that the identity sin2A+cos2=1

sin2A=1cos2sin2A=11sec2A

=sec2A1sec2A

sinA=sec2A1sec2A

=sec2A1secA

cosecA=secAsec2A1

tanA=sinAcosA=sec2A1

cotA=1sec2A1

Q 3: Choose the correct option. Justify your choice (i)9sec2A9tan2A=

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

Answer:

The correct option is (B) = 9

9sec2A9tan2A=9(sec2Atan2A) .............(i)

And it is known that sec2Atan2A=1

Therefore, equation (i) becomes, 9×1=9

Q 3: Choose the correct option. Justify your choice (ii)(1+tanθ+secθ)(1+cotθcosecθ)=

(A) 0 (B) 1 (C) 2 (D) –1

Answer:

The correct option is (C)

(1+tanθ+secθ)(1+cotθcosecθ) .......................(i)

we can write his above equation as;

=(1+sinθ/cosθ+1/cosθ)(1+cosθ/sinθ1/sinθ)=(1+sinθ+cosθ)cosθ.sinθ×((sinθ+cosθ1sinθ.cosθ)=(sinθ+cosθ)212sinθ.cosθ=sin2θ+cos2θ+2sinθ.cosθ1sinθ.cosθ=2×sinθ.cosθsinθ.cosθ = 2

Q 3: Choose the correct option. Justify your choice (iii)(secA+tanA)(1sinA)=

(A)secA (B)sinA (C)cosecA (D)cosA

Answer:

The correct option is (D)

(secA+tanA)(1sinA)=

(1cosA+sinAcosA)(1sinA)1+sinAcosA(1sinA)1sin2AcosAcosA

Q 3: Choose the correct option. Justify your choice (iv)1+tan2A1+cot2A=

(A)sec2A (B)1 (C)cot2A (D)tan2A

Answer:

The correct option is (D)

1+tan2A1+cot2A ..........................eq (i)

The above equation can be written as;

We know that cotA=1tanA

therefore,

1+tan2A1+1tan2Atan2A×(1+tan2A1+tan2A)tan2A

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (i)(cscθcotθ)2=1cosθ1+cosθ

Answer:

We need to prove-

(cscθcotθ)2=1cosθ1+cosθ

Now, taking LHS,

(cscθcotθ)2=(1sinθcosθsinθ)2

=(1cosθsinθ)2=(1cosθ)(1cosθ)sin2θ

=(1cosθ)(1cosθ)1cos2θ=(1cosθ)(1cosθ)(1cosθ)(1+cosθ)=1cosθ1+cosθ

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (ii)cosA1+sinA+1+sinAcosA=2secA

Answer:

We need to prove-

cosA1+sinA+1+sinAcosA=2secA

Taking LHS;

=cos2A+1+sin2A+2sinAcosA(1+sinA)=2(1+sinA)cosA(1+sinA)=2/cosA=2secA

= RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (iii)tanθ1cotθ+cotθ1tanθ=1+secθcscθ

[ Hint: Write the expression in terms of sinθ and cosθ ]

Answer:

We need to prove-

tanθ1cotθ+cotθ1tanθ=1+secθcosecθ

Taking LHS;

tan2θtanθ1+1tanθ(1tanθ)tan3θtan4θ+tanθ1(tanθ1).tanθ.(1tanθ)(tan3θ1)(1tanθ)tanθ.(tanθ1)(1tanθ)

By using the identity a 3 - b 3 =(a - b) (a 2 + b 2 +ab)

(tanθ1)(tan2θ+1+tanθ)tanθ(tanθ1a)tanθ+1+1tanθ1+1+tan2θtanθ1+sec2θ×1tanθ1+secθ.cscθ

= RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (iv)1+secAsecA=sin2A1cosA

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

1+secAsecA=sin2A1cosA

Taking LHS;

1+secAsecA(1+1cosA)/secA1+cosA

Taking RHS;

We know that identity 1cos2θ=sin2θ

sin2A1cosA1cos2A1cosA(1cosA)(1+cosA)(1cosA)1+cosA

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (v)cosAsinA+1cosA+sinA1=cscA+cotA , using the identity csc2A=1+cot2A

Answer:

We need to prove -

cosAsinA+1cosA+sinA1=cosecA+cotA

Dividing the numerator and denominator by sinA , we get;

=cotA1+cscAcotA+1cscA=(cotA+cscA)(csc2Acot2A)cotA+1cscA=(cscA+cotA)(1cscA+cotA)cotA+1cscA=cscA+cotA=RHS

Hence Proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (vi)1+sinA1sinA=secA+tanA

Answer:

We need to prove -

1+sinA1sinA=secA+tanA

Taking LHS;

By rationalising the denominator, we get;

=1+sinA1sinA×1+sinA1+sinA=(1+sinA)21sin2A=1+sinAcosA=secA+tanA=RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined. (vii)sinθ2sin3θ2cos3θcosθ=tanθ

Answer:

We need to prove -

sinθ2sin3θ2cos3θcosθ=tanθ

Taking LHS;

[we know the identity cos2θ=2cos2θ1=cos2θsin2θ ]

sinθ(12sin2θ)cosθ(2cos2θ1)sinθ(sin2θ+cos2θ2sinθ)cosθ.cos2θsinθ.cos2θcosθ.cos2θtanθ=RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (viii)(sinA+cscA)2+(cosA+secA)2=7+tan2A+cot2A.

Answer:

Given the equation,

(sinA+cscA)2+(cosA+secA)2=7+tan2A+cot2A ..................(i)

Taking LHS;

(sinA+cscA)2+(cosA+secA)2

sin2A+csc2A+2+cos2A+sec2A+21+2+2+(1+cot2A)+(1+tan2A)

[since sin2θ+cos2θ=1,csc2θcot2θ=1,sec2θtan2θ=1 ]

7+csc2A+tan2A=RHS

Hence proved

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (ix)(cosecAsinA)(secAcosA)=1tanA+cotA

[ Hint : Simplify LHS and RHS separately]

Answer:

We need to prove-

(coescAsinA)(secAcosA)=1tanA+cotA

Taking LHS;

(cosecA1cscA)(secA1secA)(cosec21)cosecA×sec2A1secAcot2AcosecA.tan2AsecAsinA.cosA

Taking RHS;

1sinA/cosA+cosA/sinAsinA.cosAsin2A+cos2AsinA.cosA

LHS = RHS

Hence proved.

Q 4: Prove the following identities, where the angles involved are acute angles for which the expressions are defined (x)(1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Answer:

We need to prove,

(1+tan2A1+cot2A)=(1tanA1cotA)2=tan2A

Taking LHS;

1+tan2A1+cot2A=sec2Acsc2A=tan2A

Taking RHS;

=(1tanA1cotA)2=(1sinA/cosA1cosA/sinA)2=(cosAsinA)2(sin2A)(sinAcosA)2(cos2A)=tan2A

LHS = RHS

Hence proved.


Also read-

NEET/JEE Coaching Scholarship

Get up to 90% Scholarship on Offline NEET/JEE coaching from top Institutes

JEE Main high scoring chapters and topics

As per latest 2024 syllabus. Study 40% syllabus and score upto 100% marks in JEE

Topics covered in Chapter 8 Introduction to Trigonometry: Exercise 8.3

1. Understanding basic trigonometric identities: You must learn the essential three trigonometric identities and understand their value for problem simplification and solution.

2. Verifying trigonometric identities: Perform proofs which establish two sides of an equation to be equal by using proper transformations in combination with trigonometric relations.

3. Developing logical and proof-solving skills: Step-by-step logical verification belongs to the set of skills that students need to develop for checking identities and producing mathematical proofs.

4. Preparing for advanced applications: Construct a strong foundation to solve problems utilising height-distance relationships alongside trigonometric equation methods, as well as actual angle measurement scenarios.

Check Out-

NEET/JEE Offline Coaching
Get up to 90% Scholarship on your NEET/JEE preparation from India’s Leading Coaching Institutes like Aakash, ALLEN, Sri Chaitanya & Others.
Apply Now

NCERT Solutions of Class 10 Subject Wise

Students must check the NCERT solutions for class 10 of the Mathematics and Science Subjects.

JEE Main Important Mathematics Formulas

As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters

NCERT Exemplar Solutions of Class 10 Subject-Wise

Students must check the NCERT Exemplar solutions for class 10 of the Mathematics and Science Subjects.

Frequently Asked Questions (FAQs)

1. Are the questions in Exercise 8.3 important for board exams?

Yes, Exercise 8.3 is very important as identity-based questions frequently appear in CBSE Class 10 board exams

2. Do I need to memorize all trigonometric identities?

You should at least remember the fundamental identities and standard angle values to solve Exercise 8.3 effectively.

3. Can I use a calculator for solving trigonometric identities in the exam?

No, calculators are not allowed in Class 10 board exams, so you must rely on your memorised values and identities.

Articles

Explore Top Universities Across Globe

University of Essex, Colchester
 Wivenhoe Park Colchester CO4 3SQ
University College London, London
 Gower Street, London, WC1E 6BT
The University of Edinburgh, Edinburgh
 Old College, South Bridge, Edinburgh, Post Code EH8 9YL
University of Bristol, Bristol
 Beacon House, Queens Road, Bristol, BS8 1QU
University of Nottingham, Nottingham
 University Park, Nottingham NG7 2RD

Questions related to CBSE Class 10th

Have a question related to CBSE Class 10th ?

Hello

Since you are a domicile of Karnataka and have studied under the Karnataka State Board for 11th and 12th , you are eligible for Karnataka State Quota for admission to various colleges in the state.

1. KCET (Karnataka Common Entrance Test): You must appear for the KCET exam, which is required for admission to undergraduate professional courses like engineering, medical, and other streams. Your exam score and rank will determine your eligibility for counseling.

2. Minority Income under 5 Lakh : If you are from a minority community and your family's income is below 5 lakh, you may be eligible for fee concessions or other benefits depending on the specific institution. Some colleges offer reservations or other advantages for students in this category.

3. Counseling and Seat Allocation:

After the KCET exam, you will need to participate in online counseling.

You need to select your preferred colleges and courses.

Seat allocation will be based on your rank , the availability of seats in your chosen colleges and your preferences.

4. Required Documents :

Domicile Certificate (proof that you are a resident of Karnataka).

Income Certificate (for minority category benefits).

Marksheets (11th and 12th from the Karnataka State Board).

KCET Admit Card and Scorecard.

This process will allow you to secure a seat based on your KCET performance and your category .

check link for more details

https://medicine.careers360.com/neet-college-predictor

Hope this helps you .

Hello Aspirant,  Hope your doing great,  your question was incomplete and regarding  what exam your asking.

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

View All

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

Back to top