NCERT Solutions: Introduction to Trigonometry (Exercise 8.3)

# NCERT Solutions for Class 10 Maths 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°)
So, sin 67° + cos 75° = 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°

The document NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry (Exercise 8.3) is a part of the Bank Exams Course NCERT Mathematics for Competitive Exams.
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## NCERT Mathematics for Competitive Exams

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## FAQs on NCERT Solutions for Class 10 Maths Chapter 8 - Introduction to Trigonometry (Exercise 8.3)

 1. What is trigonometry?
Ans. Trigonometry is a branch of mathematics that deals with the study of the relationships between the angles and sides of triangles. It helps in solving problems related to angles, distances, heights, and navigation.
 2. What are the basic trigonometric ratios?
Ans. The basic trigonometric ratios are sine, cosine, and tangent. They are defined as follows: - Sine (sin) = Opposite/Hypotenuse - Cosine (cos) = Adjacent/Hypotenuse - Tangent (tan) = Opposite/Adjacent
 3. How can trigonometry be applied in real-life situations?
Ans. Trigonometry has various applications in real-life situations, such as: - It is used in astronomy to calculate the distance between stars and planets. - It is used in architecture to determine the height and angles of buildings. - It is used in navigation and GPS systems to calculate distances and directions. - It is used in engineering for designing structures and calculating forces.
 4. What are the inverse trigonometric functions?
Ans. The inverse trigonometric functions are arcsine (sin⁻¹), arccosine (cos⁻¹), and arctangent (tan⁻¹). They are used to find the angle when the trigonometric ratio is known.
 5. How can I use trigonometry to find the length of a side in a right-angled triangle?
Ans. To find the length of a side in a right-angled triangle, you can use the trigonometric ratios. For example, if you know one angle and the length of one side, you can use the sine, cosine, or tangent ratio to find the length of the other side.

## NCERT Mathematics for Competitive Exams

276 docs|149 tests

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