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

NCERT solutions for class 10 Maths Chapter 8, Introduction to Trigonometry, Exercise 8.3, is available in free pdf format and it is easy to access and download. The NCERT solutions for class 10 maths consists of all solutions with proper diagrams with detailed explanation in a step by step procedure.

The solutions are created and examined by talented professors and subject matter experts to give accurate answers. As per the latest syllabus and guidelines of CBSE board, the NCERT solutions are created for the students to learn with ease. It is helpful for the students in both the board examination as well as in the competitive examinations also.

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

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Access other exercise solutions of class 10 Maths Chapter 8- Introduction to Trigonometry

Exercise 8.1 Solutions – 11 Questions (7 short answers, 3 long answers, 1 short answer with reasoning)

Exercise 8.2 Solutions – 4 Questions ( 1 short answer, 2 long answers, 1 MCQ)

Exercise 8.4 Solutions – 5 Questions ( 2 short answers, 2 long answers, 1 MCQ)

Access Answers of Maths NCERT class 10 Chapter 8 – Introduction to Trigonometry Exercise 8.3

1. Evaluate :

(i) sin 18°/cos 72°        

(ii) tan 26°/cot 64°      

(iii)  cos 48° – sin 42°      

(iv)  cosec 31° – sec 59°

Solution:

(i) sin 18°/cos 72°

To simplify this, convert the sin function into cos function

We know that, 18° is written as 90° – 18°, which is equal to the cos 72°. 

= sin (90° – 18°) /cos 72° 

Substitute the value, to simplify this equation

= cos 72° /cos 72° = 1


(ii) tan 26°/cot 64°

To simplify this, convert the tan function into cot function

We know that, 26° is written as 90° – 36°, which is equal to the cot 64°. 

= tan (90° – 36°)/cot 64°

Substitute the value, to simplify this equation

= cot 64°/cot 64° = 1


(iii) cos 48° – sin 42°

To simplify this, convert the cos function into sin function

We know that, 48° is written as 90° – 42°, which is equal to the sin 42°. 

= cos (90° – 42°) – sin 42°

Substitute the value, to simplify this equation

= sin 42° – sin 42° = 0


(iv) cosec 31° – sec 59°

To simplify this, convert the cosec function into sec function

We know that, 31° is written as 90° – 59°, which is equal to the sec 59°

 = cosec (90° – 59°) – sec 59°

Substitute the value, to simplify this equation

 = sec 59° – sec 59° = 0

2.  Show that:

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

(ii) cos 38° cos 52° – sin 38° sin 52° = 0

Solution:

(i) tan 48° tan 23° tan 42° tan 67°

Simplify the given problem by converting some of the tan functions to the cot functions

We know that, tan 48° = tan (90° – 42°) = cot 42°

tan 23° = tan (90° – 67°) = cot 67°

= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°

Substitute the values

= cot 42° cot 67° tan 42° tan 67°

= (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

(ii) cos 38° cos 52° – sin 38° sin 52°

Simplify the given problem by converting some of the cos functions to the sin functions

We know that,

cos 38° = cos (90° – 52°) = sin 52°

cos 52°= cos (90°-38°) = sin 38°

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

Substitute the values

= sin 52° sin 38° – sin 38° sin 52° = 0

3. If tan 2A = cot (A – 18°), where 2A is an acute angle, find the value of A.

Solution:

tan 2A = cot (A- 18°)

We know that tan 2A = cot (90° – 2A)

Substitute the above equation in the given problem

⇒ cot (90° – 2A) = cot (A -18°)

Now, equate the angles,

⇒ 90° – 2A = A- 18° ⇒ 108° = 3A

A = 108° / 3

Therefore, the value of A = 36°

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

Solution:

tan A = cot B

We know that cot B = tan (90° – B)

To prove A + B = 90°, substitute the above equation in the given problem

tan A = tan (90° – B)

A = 90° – B

A + B = 90°

Hence Proved.

5. If sec 4A = cosec (A – 20°), where 4A is an acute angle, find the value of A.

Solution:

sec 4A = cosec (A – 20°)

We know that sec 4A = cosec (90° – 4A)

To find the value of A, substitute the above equation in the given problem

cosec (90° – 4A) = cosec (A – 20°)

Now, equate the angles

90° – 4A= A- 20°

110° = 5A

A = 110°/ 5 = 22°

Therefore, the value of A = 22°

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

    sin (B+C/2) = cos A/2

Solution:

We know that, for a given triangle, sum of all the interior angles of a triangle is equal to 180°

A + B + C = 180° ….(1)

To find the value of (B+ C)/2, simplify the equation (1)

⇒ B + C = 180° – A

⇒ (B+C)/2 = (180°-A)/2

⇒ (B+C)/2 = (90°-A/2)

Now, multiply both sides by sin functions, we get

⇒ sin (B+C)/2 = sin (90°-A/2)

Since sin (90°-A/2) = = cos A/2, the above equation is equal to

sin (B+C)/2 = cos A/2

Hence proved.

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

Solution:

Given:

sin 67° + cos 75°

In term of sin as cos function and cos as sin function, it can be written as follows

sin 67° = sin (90° – 23°)

cos 75° = cos (90° – 15°)

= sin (90° – 23°) + cos (90° – 15°)

Now, simplify the above equation

= cos 23° + sin 15°

Therefore, sin 67° + cos 75° is also expressed as cos 23° + sin 15°


Exercise 8.3 of class 10 maths recalls the concept of complementary angles. In a right angle triangle when two angles are said to be complementary when its sum equals to 90°. In this chapter, the complementary angles of six trigonometric functions like sin, cos, tan, cosec, sec and cot are formulated with respect to the angles that lie between 0° and 90°.

In this chapter, the problem to simplify the expressions of complex trigonometric functions, finding the value of some specific angle and expressing the trigonometric function with respect to other trigonometric functions are given in exercise 8.3 and also learn the complete NCERT solutions for class 10 maths chapter 8 to score more in the final exam.

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