NCERT Solutions of Class 10 Maths Chapter 8: Introduction to Trigonometry

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NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry are given in the article. Trigonometry studies relationships between side lengths and angles of triangles. Chapter 8 Introduction to Trigonometry solutions cover key concepts such including Trigonometric ratios of an acute angle of a right-angled triangle & their proofs. Values of the trigonometric ratios of 300, 400, and 600 and relationships between Ratios.

Class 10 Maths Chapter 8 Introduction to Trigonometry is a part of Unit 5 Trigonometry. This unit holds a weightage of 12 Marks in Class 10 Maths Examination 2022-23. NCERT Solutions of Class 10 Maths Chapter 8 are based on the following topics:

Download: NCERT Solutions for Class 12 Mathametics Chapter 8 pdf


NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry

NCERT Solutions of Class 10 Maths Chapter 8 Introduction to Trigonometry are provided below: 

Also check: Introduction to Trigonometry


Important Topics: NCERT Solutions of 10 Maths Chapter 8 Introduction to Trigonometry

Important Topics of NCERT Solutions of 10 Maths Chapter 8 Introduction to Trigonometry are elaborated below:

Trigonometric Identities

There are 6 kinds of Trigonometric Idenities:

  • Sine 
  • Cosine 
  • Tangent 
  • Secant 
  • Cosecant 
  • Cotangent

Example Prove the following identity using the trigonometric identities:

[(sin 3θ + cos 3θ)/(sin θ + cos θ)] + sin θ cos θ = 1

Solution: Let’s the following identity:

a3+b3 = (a+b)(a2-ab+b2)

Using Pythagoras Theorem, 

L.H.S. = [(sin 3θ + cos 3θ)(sin θ + cos θ)] + sin θ cos θ

= [(sin θ + cos θ)(sin2θ - sin θ cos θ + cos2θ)(sin θ + cos θ) + sin θ cos θ

= (sin2θ - sin θ cos θ + cos2θ) + sin θ cos θ

= sin2θ + cos2θ = 1 = R.H.S.

Trigonometry Table

Trigonometry Table is a collection of values of trigonometric ratios for various standard angles including 0°, 30°, 45°, 60°, 90°, sometimes with other angles like 180°, 270°, and 360° included, in a tabular format.

Example: Calculate the exact value of sin15º using the trigonometric value for standard angles from a trigonometry table.

Solution: Using the trigonometric table, we know that sin45º = 1/√2, cos30º = (√3/2), cos45º = 1/√2, and sin30º = 1/2

sin15º = sin(45º - 30º) = sin45ºcos30º - cos45ºsin30º = (√2/2) • (√3/2) - (√2/2) • (1/2) = (√6 - √2)/4 = (√3 - 1)/2√2

Therefore, tThe value of sin15º = (√3 - 1)/2√2

Trigonometry Ratios

Trigonometric ratios are the “ratios of length of sides of a triangle”.

Trigonometric ratios relate the ratio of sides of a right triangle to the respective angle. Basic trigonometric ratios are sin, cos, and tan. Other important trigonometric ratios, cosec, sec, and cot, can be derived using the sin, cos, and tan respectively.

Sum, Difference, Product Trigonometric Ratios Identities

Sum, Difference, Product Trigonometric Ratios include the following formulas:

  • sin (A + B) = sin A cos B + cos A sin B
  • sin (A - B) = sin A cos B - cos A sin B
  • cos (A + B) = cos A cos B - sin A sin B
  • cos (A - B) = cos A cos B + sin A sin B
  • tan (A + B) = (tan A + tan B)/ (1 - tan A tan B)
  • tan (A - B) = (tan A - tan B)/ (1 + tan A tan B)
  • cot (A + B) = (cot A cot B - 1)/(cot B - cot A)
  • cot (A - B) = (cot A cot B + 1)/(cot B - cot A)
  • 2 sin A⋅cos B = sin(A + B) + sin(A - B)
  • 2 cos A⋅cos B = cos(A + B) + cos(A - B)
  • 2 sin A⋅sin B = cos(A - B) - cos(A + B)

NCERT Solutions For Class 10 Maths Chapter 8 Exercises

The detailed solutions for all the NCERT Solutions for Introduction to Trigonometry under different exercises are as follows:

Also check:

Also check:

CBSE X Related Questions

  • 1.
    The graph of \(y = f(x)\) is given. The number of zeroes of \(f(x)\) is :

      • 0
      • 1
      • 3
      • 2

    • 2.
      A conical cavity of maximum volume is carved out from a wooden solid hemisphere of radius 10 cm. Curved surface area of the cavity carved out is (use \(\pi = 3.14\))

        • \(314 \sqrt{2}\) \(\text{cm}^{2}\)
        • \(314\) \(\text{cm}^{2}\)
        • \(\frac{3140}{3}\) \(\text{cm}^{2}\)
        • \(3140 \sqrt{2}\) \(\text{cm}^{2}\)

      • 3.
        PQ is tangent to a circle with centre O. If \(OQ = a\), \(OP = a + 2\) and \(PQ = 2b\), then relation between \(a\) and \(b\) is

          • \(a^2 + (a + 2)^2 = (2b)^2\)
          • \(b^2 = a + 4\)
          • \(2a^2 + 1 = b^2\)
          • \(b^2 = a + 1\)

        • 4.
          Prove that: \(\frac{\sec^3 \theta}{\sec^2 \theta - 1} + \frac{\csc^3 \theta}{\csc^2 \theta - 1} = \sec \theta \cdot \csc \theta (\sec \theta + \csc \theta)\)


            • 5.
              Assertion (A) : \((\sqrt{3} + \sqrt{5})\) is an irrational number.
              Reason (R) : Sum of the any two irrational numbers is always irrational.

                • Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
                • Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of Assertion (A).
                • Assertion (A) is true, but Reason (R) is false.
                • Assertion (A) is false, but Reason (R) is true.

              • 6.
                An ice-cream cone of radius \(r\) and height \(h\) is completely filled by two spherical scoops of ice-cream. If radius of each spherical scoop is \(\frac{r}{2}\), then \(h : 2r\) equals

                  • \(1 : 8\)
                  • \(1 : 2\)
                  • \(1 : 1\)
                  • \(2 : 1\)

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