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Samacheer Kalvi 10th Maths Exercise 6.5 Solutions

10th Maths Exercise 6.5 solutions (Chapter 6 Trigonometry) with step-by-step answers for all 15 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance15 answers · full steps below
  1. Q1
    1. 11
  2. Q2
    1. cot⁡θ\cot\theta
  3. Q3
    1. 77
  4. Q4
    1. 2a2a
  5. Q5
    1. 125\dfrac{1}{25}
  6. Q6
    1. 32\dfrac{3}{2}
  7. Q7
    1. y2b2−x2a2=1\dfrac{y^2}{b^2} - \dfrac{x^2}{a^2} = 1
  8. Q8
    1. 22
  9. Q9
    1. b2−a2b^2 - a^2
  10. Q10
    1. 60∘60^\circ
  11. Q11
    1. b3\dfrac{b}{3}
  12. Q12
    1. 43.92 m
  13. Q13
    1. 30, 10330,\ 10\sqrt{3}
  14. Q14
    1. x22\dfrac{x}{2\sqrt{2}}
  15. Q15
    1. h(1+tan⁡β)1−tan⁡β\dfrac{h(1+\tan\beta)}{1-\tan\beta}

Q1. 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

  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.

Answer
  1. 11

Q2. 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

  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.

Answer
  1. cot⁡θ\cot\theta

Q3. 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

  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.

Answer
  1. 77

Q4. 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 …

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

  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.

Answer
  1. 2a2a

Q5. 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

  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}.

Answer
  1. 125\dfrac{1}{25}

Q6. 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}

  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}.

Answer
  1. 32\dfrac{3}{2}

Q7. 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

  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.

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

Q8. (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

  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.

Answer
  1. 22

Q9. 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

  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.

Answer
  1. b2−a2b^2 - a^2

Q10. If the ratio of the height of a tower and the length …

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

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

  2. θ=60∘\theta=60^\circ.

Answer
  1. 60∘60^\circ

Q11. The electric pole subtends an angle of 30∘30^\circ at a point on …

The electric pole subtends an angle of 30∘30^\circ at a point on the same level as its foot. At a second point 'bb' metres above the first, the depression of the foot of the pole is 60∘60^\circ. The height of the pole (in metres) is equal to

  1. (1) 3 b\sqrt{3}\,b
  2. (2) b3\dfrac{b}{3}
  3. (3) b2\dfrac{b}{2}
  4. (4) b3\dfrac{b}{\sqrt{3}}

  1. Let dd be the horizontal distance from the pole and hh its height.

  2. From the point bb m high, depression of the foot is 60∘60^\circ: tan⁡60∘=bd⇒d=b3\tan 60^\circ=\dfrac{b}{d}\Rightarrow d=\dfrac{b}{\sqrt3}.

  3. From the first point, elevation of the top is 30∘30^\circ: tan⁡30∘=hd\tan 30^\circ=\dfrac{h}{d}.

  4. h=13×b3=b3h=\dfrac{1}{\sqrt3}\times\dfrac{b}{\sqrt3}=\dfrac{b}{3}.

Answer
  1. b3\dfrac{b}{3}

Q12. A tower is 60 m heigh. Its shadow reduces by xx metres …

A tower is 60 m heigh. Its shadow reduces by xx metres when the angle of elevation of the sun increases from 30∘30^\circ to 45∘45^\circ then xx is equal to

  1. (1) 41.92 m
  2. (2) 43.92 m
  3. (3) 43 m
  4. (4) 45.6 m

  1. Shadow at 30∘30^\circ: 60tan⁡30∘=603\dfrac{60}{\tan 30^\circ}=60\sqrt3 m.

  2. Shadow at 45∘45^\circ: 60tan⁡45∘=60\dfrac{60}{\tan 45^\circ}=60 m.

  3. x=603−60=60(1.732−1)=43.92x=60\sqrt3-60=60(1.732-1)=43.92 m.

Answer
  1. 43.92 m

Q13. The angle of depression of the top and bottom of 20 m …

The angle of depression of the top and bottom of 20 m tall building from the top of a multistoried building are 30∘30^\circ and 60∘60^\circ respectively. The height of the multistoried building and the distance between two buildings (in metres) is

  1. (1) 20, 10320,\ 10\sqrt{3}
  2. (2) 30, 5330,\ 5\sqrt{3}
  3. (3) 20, 1020,\ 10
  4. (4) 30, 10330,\ 10\sqrt{3}

  1. Let hh be the height of the multistoried building and dd the distance between them.

  2. tan⁡60∘=hd⇒h=d3\tan 60^\circ=\dfrac{h}{d}\Rightarrow h=d\sqrt3.

  3. tan⁡30∘=h−20d⇒h−20=d3\tan 30^\circ=\dfrac{h-20}{d}\Rightarrow h-20=\dfrac{d}{\sqrt3}.

  4. d3−d3=20⇒2d3=20⇒d=103d\sqrt3-\dfrac{d}{\sqrt3}=20\Rightarrow \dfrac{2d}{\sqrt3}=20\Rightarrow d=10\sqrt3.

  5. h=103×3=30h=10\sqrt3\times\sqrt3=30 m.

Answer
  1. 30, 10330,\ 10\sqrt{3}

Q14. Two persons are standing 'xx' metres apart from each other and the …

Two persons are standing 'xx' metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is

  1. (1) 2 x\sqrt{2}\,x
  2. (2) x22\dfrac{x}{2\sqrt{2}}
  3. (3) x2\dfrac{x}{\sqrt{2}}
  4. (4) 2x2x

  1. Let the shorter height be hh; the other is 2h2h. The observer is at the midpoint, x2\dfrac{x}{2} from each.

  2. tan⁡θ=2hx/2=4hx\tan\theta=\dfrac{2h}{x/2}=\dfrac{4h}{x} and tan⁡(90∘−θ)=cot⁡θ=hx/2=2hx\tan(90^\circ-\theta)=\cot\theta=\dfrac{h}{x/2}=\dfrac{2h}{x}.

  3. tan⁡θ⋅cot⁡θ=1⇒4hx×2hx=1\tan\theta\cdot\cot\theta=1\Rightarrow\dfrac{4h}{x}\times\dfrac{2h}{x}=1.

  4. 8h2=x2⇒h=x228h^2=x^2\Rightarrow h=\dfrac{x}{2\sqrt2}.

Answer
  1. x22\dfrac{x}{2\sqrt{2}}

Q15. The angle of elevation of a cloud from a point hh metres …

The angle of elevation of a cloud from a point hh metres above a lake is β\beta. The angle of depression of its reflection in the lake is 45∘45^\circ. The height of location of the cloud from the lake is

  1. (1) h(1+tan⁡β)1−tan⁡β\dfrac{h(1+\tan\beta)}{1-\tan\beta}
  2. (2) h(1−tan⁡β)1+tan⁡β\dfrac{h(1-\tan\beta)}{1+\tan\beta}
  3. (3) htan⁡(45∘−β)h\tan(45^\circ-\beta)
  4. (4) none of these

  1. Let the cloud be at height HH above the lake and dd the horizontal distance.

  2. Elevation: tan⁡β=H−hd\tan\beta=\dfrac{H-h}{d}; depression of reflection: tan⁡45∘=H+hd\tan 45^\circ=\dfrac{H+h}{d}.

  3. So d=H+hd=H+h and (H−h)=(H+h)tan⁡β(H-h)=(H+h)\tan\beta.

  4. H(1−tan⁡β)=h(1+tan⁡β)⇒H=h(1+tan⁡β)1−tan⁡βH(1-\tan\beta)=h(1+\tan\beta)\Rightarrow H=\dfrac{h(1+\tan\beta)}{1-\tan\beta}.

Answer
  1. h(1+tan⁡β)1−tan⁡β\dfrac{h(1+\tan\beta)}{1-\tan\beta}
Chapter 6 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}

Answers checked against the Samacheer Kalvi textbook · Updated

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