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10th Maths Chapter 6 Trigonometry

10th Maths Chapter 6 Trigonometry solutions, with the chapter's formulas and every exercise worked step by step, as per the 2026-27 Samacheer Kalvi textbook.

This chapter builds on the basic trigonometric ratios and introduces the three standard identities. Students learn to prove that an equation holds for every value of the angle and to simplify expressions using these identities. The chapter then applies trigonometry to real situations through the line of sight, angles of elevation and depression. Exercises ask students to prove identities and to find heights of towers, buildings and distances between objects.

Exercises in this chapter

Key formulas

  • Pythagorean identity

    sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

  • Identity with tangent and secant

    1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta

  • Identity with cotangent and cosecant

    1+cot⁡2θ=cosec⁡2θ1 + \cot^2\theta = \operatorname{cosec}^2\theta

  • Trigonometric identity (definition)

    An equation in trigonometric ratios that is true for all values of the angle\text{An equation in trigonometric ratios that is true for all values of the angle}

  • Line of sight

    The line drawn from the eye of an observer to the point on the object being viewed\text{The line drawn from the eye of an observer to the point on the object being viewed}

  • Angle of elevation

    Angle between the line of sight and the horizontal, when the object is above the horizontal level\text{Angle between the line of sight and the horizontal, when the object is above the horizontal level}

  • Angle of depression

    Angle between the line of sight and the horizontal, when the object is below the horizontal level\text{Angle between the line of sight and the horizontal, when the object is below the horizontal level}

  • Use of trigonometric ratios

    Height or length of an object, or distance between two far objects, can be found using trigonometric ratios\text{Height or length of an object, or distance between two far objects, can be found using trigonometric ratios}

Chapter 6 one-mark practice

  1. The value of sin⁡2θ+11+tan⁡2θ\sin^2\theta + \dfrac{1}{1+\tan^2\theta} is equal to

    1. (1) tan⁡2θ\tan^2\theta
    2. (2) 11
    3. (3) cot⁡2θ\cot^2\theta
    4. (4) 00
    Show answer

    Answer: (2) 11

    1. 1+tan⁡2θ=sec⁡2θ1+\tan^2\theta=\sec^2\theta, so 11+tan⁡2θ=cos⁡2θ\dfrac{1}{1+\tan^2\theta}=\cos^2\theta.

    2. sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1.

    Question 1 of 10

  2. tan⁡θ cosec2θ−tan⁡θ\tan\theta\,\mathrm{cosec}^2\theta - \tan\theta is equal to

    1. (1) sec⁡θ\sec\theta
    2. (2) cot⁡2θ\cot^2\theta
    3. (3) sin⁡θ\sin\theta
    4. (4) cot⁡θ\cot\theta
    Show answer

    Answer: (4) cot⁡θ\cot\theta

    1. tan⁡θ cosec2θ−tan⁡θ=tan⁡θ(cosec2θ−1)\tan\theta\,\mathrm{cosec}^2\theta-\tan\theta=\tan\theta(\mathrm{cosec}^2\theta-1).

    2. cosec2θ−1=cot⁡2θ\mathrm{cosec}^2\theta-1=\cot^2\theta.

    3. =tan⁡θ⋅cot⁡2θ=cot⁡θ=\tan\theta\cdot\cot^2\theta=\cot\theta.

    Question 2 of 10

  3. If (sin⁡α+cosec α)2+(cos⁡α+sec⁡α)2=k+tan⁡2α+cot⁡2α(\sin\alpha + \mathrm{cosec}\,\alpha)^2 + (\cos\alpha + \sec\alpha)^2 = k + \tan^2\alpha + \cot^2\alpha, then the value of kk is equal to

    1. (1) 99
    2. (2) 77
    3. (3) 55
    4. (4) 33
    Show answer

    Answer: (2) 77

    1. (sin⁡α+cosec α)2=sin⁡2α+2+cosec2α(\sin\alpha+\mathrm{cosec}\,\alpha)^2=\sin^2\alpha+2+\mathrm{cosec}^2\alpha.

    2. (cos⁡α+sec⁡α)2=cos⁡2α+2+sec⁡2α(\cos\alpha+\sec\alpha)^2=\cos^2\alpha+2+\sec^2\alpha.

    3. Sum =(sin⁡2α+cos⁡2α)+4+(1+cot⁡2α)+(1+tan⁡2α)=(\sin^2\alpha+\cos^2\alpha)+4+(1+\cot^2\alpha)+(1+\tan^2\alpha).

    4. =7+tan⁡2α+cot⁡2α=7+\tan^2\alpha+\cot^2\alpha, so k=7k=7.

    Question 3 of 10

  4. If sin⁡θ+cos⁡θ=a\sin\theta + \cos\theta = a and sec⁡θ+cosec θ=b\sec\theta + \mathrm{cosec}\,\theta = b, then the value of b(a2−1)b(a^2-1) is equal to

    1. (1) 2a2a
    2. (2) 3a3a
    3. (3) 00
    4. (4) 2ab2ab
    Show answer

    Answer: (1) 2a2a

    1. a2−1=(sin⁡θ+cos⁡θ)2−1=2sin⁡θcos⁡θa^2-1=(\sin\theta+\cos\theta)^2-1=2\sin\theta\cos\theta.

    2. b=sec⁡θ+cosec θ=sin⁡θ+cos⁡θsin⁡θcos⁡θ=asin⁡θcos⁡θb=\sec\theta+\mathrm{cosec}\,\theta=\dfrac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\dfrac{a}{\sin\theta\cos\theta}.

    3. b(a2−1)=asin⁡θcos⁡θ×2sin⁡θcos⁡θ=2ab(a^2-1)=\dfrac{a}{\sin\theta\cos\theta}\times 2\sin\theta\cos\theta=2a.

    Question 4 of 10

  5. If 5x=sec⁡θ5x = \sec\theta and 5y=tan⁡θ\dfrac{5}{y} = \tan\theta, then x2−1y2x^2 - \dfrac{1}{y^2} is equal to

    1. (1) 2525
    2. (2) 125\dfrac{1}{25}
    3. (3) 55
    4. (4) 11
    Show answer

    Answer: (2) 125\dfrac{1}{25}

    1. x=sec⁡θ5x=\dfrac{\sec\theta}{5} and 1y=tan⁡θ5\dfrac{1}{y}=\dfrac{\tan\theta}{5}.

    2. x2−1y2=sec⁡2θ−tan⁡2θ25=125x^2-\dfrac{1}{y^2}=\dfrac{\sec^2\theta-\tan^2\theta}{25}=\dfrac{1}{25}.

    Question 5 of 10

  6. If sin⁡θ=cos⁡θ\sin\theta = \cos\theta, then 2tan⁡2θ+sin⁡2θ−12\tan^2\theta + \sin^2\theta - 1 is equal to

    1. (1) −32\dfrac{-3}{2}
    2. (2) 32\dfrac{3}{2}
    3. (3) 23\dfrac{2}{3}
    4. (4) −23\dfrac{-2}{3}
    Show answer

    Answer: (2) 32\dfrac{3}{2}

    1. sin⁡θ=cos⁡θ⇒tan⁡θ=1⇒θ=45∘\sin\theta=\cos\theta\Rightarrow\tan\theta=1\Rightarrow\theta=45^\circ.

    2. 2tan⁡245∘+sin⁡245∘−1=2(1)+12−1=322\tan^2 45^\circ+\sin^2 45^\circ-1=2(1)+\dfrac{1}{2}-1=\dfrac{3}{2}.

    Question 6 of 10

  7. If x=atan⁡θx = a\tan\theta and y=bsec⁡θy = b\sec\theta then

    1. (1) y2b2−x2a2=1\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1
    2. (2) x2a2−y2b2=1\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1
    3. (3) x2a2+y2b2=1\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1
    4. (4) x2a2−y2b2=0\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 0
    Show answer

    Answer: (1) y2b2−x2a2=1\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1

    1. xa=tan⁡θ\dfrac{x}{a}=\tan\theta, yb=sec⁡θ\dfrac{y}{b}=\sec\theta.

    2. sec⁡2θ−tan⁡2θ=1⇒y2b2−x2a2=1\sec^2\theta-\tan^2\theta=1\Rightarrow\dfrac{y^2}{b^2}-\dfrac{x^2}{a^2}=1.

    Question 7 of 10

  8. (1+tan⁡θ+sec⁡θ)(1+cot⁡θ−cosec θ)(1+\tan\theta+\sec\theta)(1+\cot\theta-\mathrm{cosec}\,\theta) is equal to

    1. (1) 00
    2. (2) 11
    3. (3) 22
    4. (4) −1-1
    Show answer

    Answer: (3) 22

    1. Write the second bracket as 1+cos⁡θsin⁡θ−1sin⁡θ=sin⁡θ+cos⁡θ−1sin⁡θ1+\dfrac{\cos\theta}{\sin\theta}-\dfrac{1}{\sin\theta}=\dfrac{\sin\theta+\cos\theta-1}{\sin\theta}.

    2. First bracket =cos⁡θ+sin⁡θ+1cos⁡θ=\dfrac{\cos\theta+\sin\theta+1}{\cos\theta}.

    3. Product =(sin⁡θ+cos⁡θ)2−1sin⁡θcos⁡θ=2sin⁡θcos⁡θsin⁡θcos⁡θ=2=\dfrac{(\sin\theta+\cos\theta)^2-1}{\sin\theta\cos\theta}=\dfrac{2\sin\theta\cos\theta}{\sin\theta\cos\theta}=2.

    Question 8 of 10

  9. acot⁡θ+b cosec θ=pa\cot\theta + b\,\mathrm{cosec}\,\theta = p and bcot⁡θ+a cosec θ=qb\cot\theta + a\,\mathrm{cosec}\,\theta = q then p2−q2p^2 - q^2 is equal to

    1. (1) a2−b2a^2 - b^2
    2. (2) b2−a2b^2 - a^2
    3. (3) a2+b2a^2 + b^2
    4. (4) b−ab - a
    Show answer

    Answer: (2) b2−a2b^2 - a^2

    1. p2−q2=(acot⁡θ+b cosec θ)2−(bcot⁡θ+a cosec θ)2p^2-q^2=(a\cot\theta+b\,\mathrm{cosec}\,\theta)^2-(b\cot\theta+a\,\mathrm{cosec}\,\theta)^2.

    2. =(a2−b2)cot⁡2θ−(a2−b2)cosec2θ=(a^2-b^2)\cot^2\theta-(a^2-b^2)\mathrm{cosec}^2\theta (the 2ab2ab terms cancel).

    3. =(a2−b2)(cot⁡2θ−cosec2θ)=(a2−b2)(−1)=b2−a2=(a^2-b^2)(\cot^2\theta-\mathrm{cosec}^2\theta)=(a^2-b^2)(-1)=b^2-a^2.

    Question 9 of 10

  10. If the ratio of the height of a tower and the length of its shadow is 3:1\sqrt{3}:1, then the angle of elevation of the sun has measure

    1. (1) 45∘45^\circ
    2. (2) 30∘30^\circ
    3. (3) 90∘90^\circ
    4. (4) 60∘60^\circ
    Show answer

    Answer: (4) 60∘60^\circ

    1. tan⁡θ=heightshadow=31=3\tan\theta=\dfrac{\text{height}}{\text{shadow}}=\dfrac{\sqrt3}{1}=\sqrt3.

    2. θ=60∘\theta=60^\circ.

    Question 10 of 10

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