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

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

Updated on Nov 27, 2023 09:11 AM IST | #CBSE Class 10th

NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.3

NCERT Solutions for Exercise 8.3 Class 10 Maths Chapter 8 Introduction to Trigonometry are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. Class 10 maths ex 8.3 gives a brief overview of the trigonometric ratios of complementary angles and is explained in trigonometry. If the sum of two angles equals 90°, they are said to be complementary. According to the angle sum property, one angle in a right-angled triangle measures 90°, and the sum of the other two angles also measures 90°. This is the concept from which the formulas for this exercise are derived or the formula of complementary angle. The majority of these questions require proof. As a result, students must carefully study the theory in order to provide detailed answers to these sums.

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  1. NCERT Solutions For Class 10 Maths Chapter 8 Exercise 8.3
  2. Download Free Pdf of NCERT Solutions for Class 10 Maths chapter 8 exercise 8.3
  3. Assess NCERT Solutions for Class 10 Maths chapter 8 exercise 8.3
  4. More About NCERT Solutions for Class 10 Maths Exercise 8.3
  5. Benefits of NCERT Solutions for Class 10 Maths Exercise 8.3
  6. NCERT Solutions Subject Wise
  7. Subject wise NCERT Exemplar solutions
NCERT Solutions for Exercise 8.3 Class 10 Maths Chapter 8 - Introduction to Trigonometry
NCERT Solutions for Exercise 8.3 Class 10 Maths Chapter 8 - Introduction to Trigonometry

10th class Maths exercise 8.3 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 10 Maths chapter 8 exercise 8.3

Introduction to Trigonometry class 10 chapter 8 Exercise: 8.3

Q1 Evaluate :

(i)sin⁡18ocos⁡72o

Answer:

sin⁡18ocos⁡72o
We can write the above equation as;
=sin⁡(900−720)cos⁡720
By using the identity of sin⁡(90o−θ)=cos⁡θ
Therefore, cos⁡720cos⁡720=1

So, the answer is 1.

Q1 Evaluate :

(ii)tan⁡26ocot⁡64o

Answer:

tan⁡26ocot⁡64o
The above equation can be written as ;

tan⁡(90o−64o)/cot⁡64o .........(i)
It is known that, tan⁡(90o−θ)=cot⁡θ
Therefore, equation (i) becomes,
cot⁡64o/cot⁡64o=1

So, the answer is 1.

Q1 Evaluate :

(iii)cos⁡48o−sin⁡42o

Answer:

cos⁡48o−sin⁡42o
The above equation can be written as ;
cos⁡(90o−42o)−sin⁡42o ....................(i)
It is known that cos⁡(90o−θ)=sin⁡θ
Therefore, equation (i) becomes,
sin⁡42o−sin⁡42o=0

So, the answer is 0.

Q1 Evaluate :

(iv)cosec31o−sec⁡59o

Answer:

cosec31o−sec⁡59o

This equation can be written as;
cosec31o−sec⁡(90o−31o) .................(i)
We know that sec⁡(90o−θ)=cosecθ

Therefore, equation (i) becomes;
cosec31o−cosec 31o = 0

So, the answer is 0.

Q2 Show that :

(i)tan⁡48otan⁡23otan⁡42otan⁡67o=1

Answer:

tan⁡48otan⁡23otan⁡42otan⁡67o=1
Taking Left Hand Side (LHS)
= tan⁡48otan⁡23otan⁡42otan⁡67o
⇒tan⁡48otan⁡23otan⁡(90o−48o)tan⁡(90o−23o)
⇒tan⁡48otan⁡23ocot⁡48ocot⁡23o [it is known that tan⁡(900−θ=cot⁡θ) and cot⁡θ×tan⁡θ=1
=1

Hence proved.

Q2 Show that :

(ii)cos⁡38ocos⁡52o−sin⁡38osin⁡52o=0

Answer:

cos⁡38ocos⁡52o−sin⁡38osin⁡52o=0

Taking Left Hand Side (LHS)
= cos⁡38ocos⁡52o−sin⁡38osin⁡52o
= cos⁡38ocos⁡(90o−38o)−sin⁡38osin⁡(90o−38o)
= cos⁡38osin⁡38o−sin⁡38ocos⁡38o [it is known that sin⁡(900−θ)=cos⁡θ and cos⁡(900−θ)=sin⁡θ ]
= 0

Q3 If tan⁡2A=cot⁡(A−18o) , where 2A is an acute angle, find the value of A .

Answer:

We have,
tan 2A = cot (A - \180 )
we know that, cot⁡(900−θ)=tan⁡θ
⇒cot⁡(900−2A)=cot⁡(A−180)⇒900−2A=A−180⇒3A=108⇒A=108/3=360

Q4 If tan⁡A=cot⁡B , prove that A+B=90o .

Answer:

We have,
tan⁡A=cot⁡B
and we know that tan⁡(900−θ)=cot⁡θ
therefore,
tan⁡A=tan⁡(900−B)
A = 90 - B
A + B = 90
Hence proved.

Q5 If sec⁡4A=cosec(A−20o) , where 4A is an acute angle, find the value of A .

Answer:

We have,
sec⁡4A=cosec(A−20o) , Here 4A is an acute angle
According to question,
We know that cosec(900−θ)=sec⁡θ
cosec(900−4A)=cosec(A−20o)

⇒90−4A=A−20⇒5A=110⇒A=1105⇒A=22o

Q6 If A,B and C are interior angles of a triangle ABC , then show that

sin⁡(B+C2)=cos⁡A2

Answer:

Given that,
A, B and C are interior angles of ΔABC
To prove - sin⁡(B+C2)=cos⁡A2

Now,
In triangle ΔABC ,
A + B + C = 1800
⇒B+C=180−A
⇒B+C/2=900−A/2
sin⁡B+C2=sin⁡(900−A/2)
sin⁡B+C2=cos⁡A/2
Hence proved.

Q7 Express sin67o+cos⁡75o in terms of trigonometric ratios of angles between 0o and 45o .

Answer:

By using the identity of sin⁡θ and cos⁡θ
sin67o+cos⁡75o
We know that,
sin⁡(90−θ)=cos⁡θ and cos⁡(90−θ)=sin⁡θ
the above equation can be written as;
=sin⁡(900−230)+cos⁡(900−150)
=sin⁡(150)+cos⁡(230)

More About NCERT Solutions for Class 10 Maths Exercise 8.3

NCERT solutions for Class 10 Maths exercise 8.3- We can find the value of these complementary angles but in the examination, we have very little time. So, we should try to remember these formulas NCERT solutions for Class 10 Maths chapter 8 exercise 8.3 as these can break a large complex trigonometric equation into simpler ones that can be called easily by the other associated terms present in the function.

The formulas are:

sin (90 - A) = cos A

cos (90 - A) = sin A

tan (90 - A) = cot A

cot (90 - A) = tan A

sec (90 - A) = cosec A

cosec (90 - A) = sec A

for all values of angle, A lying between 0° and 90°.

We should always be careful about

tan (0) = 0 = cot (90)

sec (0) = 1 cosec (90) and sec (90), cosec(0), tan(90) and cot (0) are not defined. Students can use these Introduction to Trigonometry Class 10 notes to quick revision of the important concepts discussed in this chapter.

Benefits of NCERT Solutions for Class 10 Maths Exercise 8.3

  • Exercise 8.3 Class 10 Maths, is based on the main concept of Trigonometric Ratios of Complementary Angles.
  • Class 10 Maths chapter 8 exercise 8.3 helps in solving and revising all questions of the previous exercises. It shows us two different aspects of solving the same problem by using complementary angels.
  • Mastering the complementary angle formulas of Class 10 Maths chapter 8 exercise 8.3 can help in simplifying complex trigonometric questions into simpler ones.
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Frequently Asked Questions (FAQs)

1. In the identity sin(90-A)=cosA what is the range of angel A

Angel ‘A’ should be in between 0° and 90°.

2. if tanA=cotBthen find the value of A+B?

A+B=90 degrees

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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.

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