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Samacheer Kalvi 10th Maths Unit Exercise 6 Solutions

10th Maths Unit Exercise 6 solutions (Chapter 6 Trigonometry) with step-by-step answers for all 13 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance8 answers · full steps below
  1. Q5Speed of the bird =29.28=29.28 m/s
  2. Q6About 1.971.97 seconds (nearly 22 seconds)
  3. Q7 (i)BB is 24.57624.576 km north of AA
  4. Q7 (ii)BB is 17.20817.208 km west of AA
  5. Q7 (iii)CC is 21.4121.41 km north of BB
  6. Q7 (iv)CC is 23.7823.78 km east of BB
  7. Q8Height of the lighthouse =200=200 m
  8. Q9Height of the statue =39.19=39.19 m (approximately)

Question 1

(i)

Prove that cot⁡2A(sec⁡A−11+sin⁡A)+sec⁡2A(sin⁡A−11+sec⁡A)=0\cot^2 A\left(\dfrac{\sec A-1}{1+\sin A}\right)+\sec^2 A\left(\dfrac{\sin A-1}{1+\sec A}\right)=0

Show steps (11)

  1. Write everything in sin⁡A\sin A and cos⁡A\cos A.

  2. First term: cot⁡2A=cos⁡2Asin⁡2A\cot^2A=\dfrac{\cos^2A}{\sin^2A} and sec⁡A−1=1−cos⁡Acos⁡A\sec A-1=\dfrac{1-\cos A}{\cos A}.

  3. cot⁡2A(sec⁡A−11+sin⁡A)=cos⁡2A(1−cos⁡A)sin⁡2A cos⁡A(1+sin⁡A)=cos⁡A(1−cos⁡A)sin⁡2A(1+sin⁡A)\cot^2A\left(\frac{\sec A-1}{1+\sin A}\right)=\frac{\cos^2A(1-\cos A)}{\sin^2A\,\cos A(1+\sin A)}=\frac{\cos A(1-\cos A)}{\sin^2A(1+\sin A)}

  4. Since sin⁡2A=1−cos⁡2A=(1−cos⁡A)(1+cos⁡A)\sin^2A=1-\cos^2A=(1-\cos A)(1+\cos A),

  5. cot⁡2A(sec⁡A−11+sin⁡A)=cos⁡A(1+cos⁡A)(1+sin⁡A)\cot^2A\left(\frac{\sec A-1}{1+\sin A}\right)=\frac{\cos A}{(1+\cos A)(1+\sin A)}

  6. Second term: sec⁡2A=1cos⁡2A\sec^2A=\dfrac{1}{\cos^2A} and 1+sec⁡A=cos⁡A+1cos⁡A1+\sec A=\dfrac{\cos A+1}{\cos A}.

  7. sec⁡2A(sin⁡A−11+sec⁡A)=(sin⁡A−1)cos⁡Acos⁡2A(1+cos⁡A)=sin⁡A−1cos⁡A(1+cos⁡A)\sec^2A\left(\frac{\sin A-1}{1+\sec A}\right)=\frac{(\sin A-1)\cos A}{\cos^2A(1+\cos A)}=\frac{\sin A-1}{\cos A(1+\cos A)}

  8. Multiply numerator and denominator by (1+sin⁡A)(1+\sin A):

  9. =sin⁡2A−1cos⁡A(1+cos⁡A)(1+sin⁡A)=−cos⁡2Acos⁡A(1+cos⁡A)(1+sin⁡A)=−cos⁡A(1+cos⁡A)(1+sin⁡A)=\frac{\sin^2A-1}{\cos A(1+\cos A)(1+\sin A)}=\frac{-\cos^2A}{\cos A(1+\cos A)(1+\sin A)}=\frac{-\cos A}{(1+\cos A)(1+\sin A)}

  10. Adding the two terms:

  11. LHS=cos⁡A(1+cos⁡A)(1+sin⁡A)−cos⁡A(1+cos⁡A)(1+sin⁡A)=0=RHS\text{LHS}=\frac{\cos A}{(1+\cos A)(1+\sin A)}-\frac{\cos A}{(1+\cos A)(1+\sin A)}=0=\text{RHS}

Answer

cot⁡2A(sec⁡A−11+sin⁡A)+sec⁡2A(sin⁡A−11+sec⁡A)=0\cot^2 A\left(\dfrac{\sec A-1}{1+\sin A}\right)+\sec^2 A\left(\dfrac{\sin A-1}{1+\sec A}\right)=0, hence proved.

(ii)

Prove that tan⁡2θ−1tan⁡2θ+1=1−2cos⁡2θ\dfrac{\tan^2\theta-1}{\tan^2\theta+1}=1-2\cos^2\theta

  1. Use tan⁡2θ+1=sec⁡2θ\tan^2\theta+1=\sec^2\theta in the denominator.

  2. LHS=tan⁡2θ−1sec⁡2θ=(tan⁡2θ−1)cos⁡2θ\text{LHS}=\frac{\tan^2\theta-1}{\sec^2\theta}=(\tan^2\theta-1)\cos^2\theta

  3. =(sin⁡2θcos⁡2θ−1)cos⁡2θ=sin⁡2θ−cos⁡2θ=\left(\frac{\sin^2\theta}{\cos^2\theta}-1\right)\cos^2\theta=\sin^2\theta-\cos^2\theta

  4. Put sin⁡2θ=1−cos⁡2θ\sin^2\theta=1-\cos^2\theta:

  5. =1−cos⁡2θ−cos⁡2θ=1−2cos⁡2θ=RHS=1-\cos^2\theta-\cos^2\theta=1-2\cos^2\theta=\text{RHS}

Answer

tan⁡2θ−1tan⁡2θ+1=1−2cos⁡2θ\dfrac{\tan^2\theta-1}{\tan^2\theta+1}=1-2\cos^2\theta, hence proved.

Q2. Prove that (1+sin⁡θ−cos⁡θ1+sin⁡θ+cos⁡θ)2=1−cos⁡θ1+cos⁡θ\left(\dfrac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\right)^2=\dfrac{1-\cos\theta}{1+\cos\theta}

Show steps (8)

  1. Let N=1+sin⁡θ−cos⁡θN=1+\sin\theta-\cos\theta and D=1+sin⁡θ+cos⁡θD=1+\sin\theta+\cos\theta.

  2. ND=(1+sin⁡θ)2−cos⁡2θ=1+2sin⁡θ+sin⁡2θ−cos⁡2θND=(1+\sin\theta)^2-\cos^2\theta=1+2\sin\theta+\sin^2\theta-\cos^2\theta

  3. Using cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta:   ND=2sin⁡θ+2sin⁡2θ=2sin⁡θ(1+sin⁡θ)\;ND=2\sin\theta+2\sin^2\theta=2\sin\theta(1+\sin\theta).

  4. N2=(1+sin⁡θ)2−2cos⁡θ(1+sin⁡θ)+cos⁡2θN^2=(1+\sin\theta)^2-2\cos\theta(1+\sin\theta)+\cos^2\theta

  5. =2+2sin⁡θ−2cos⁡θ−2sin⁡θcos⁡θ=2(1+sin⁡θ)(1−cos⁡θ)=2+2\sin\theta-2\cos\theta-2\sin\theta\cos\theta=2(1+\sin\theta)(1-\cos\theta)

  6. Therefore ND=N2ND=2(1+sin⁡θ)(1−cos⁡θ)2sin⁡θ(1+sin⁡θ)=1−cos⁡θsin⁡θ\frac{N}{D}=\frac{N^2}{ND}=\frac{2(1+\sin\theta)(1-\cos\theta)}{2\sin\theta(1+\sin\theta)}=\frac{1-\cos\theta}{\sin\theta}

  7. Squaring, and using sin⁡2θ=(1−cos⁡θ)(1+cos⁡θ)\sin^2\theta=(1-\cos\theta)(1+\cos\theta):

  8. (ND)2=(1−cos⁡θ)2(1−cos⁡θ)(1+cos⁡θ)=1−cos⁡θ1+cos⁡θ\left(\frac{N}{D}\right)^2=\frac{(1-\cos\theta)^2}{(1-\cos\theta)(1+\cos\theta)}=\frac{1-\cos\theta}{1+\cos\theta}

Answer

(1+sin⁡θ−cos⁡θ1+sin⁡θ+cos⁡θ)2=1−cos⁡θ1+cos⁡θ\left(\dfrac{1+\sin\theta-\cos\theta}{1+\sin\theta+\cos\theta}\right)^2=\dfrac{1-\cos\theta}{1+\cos\theta}, hence proved.

Q3. If xsin⁡3θ+ycos⁡3θ=sin⁡θcos⁡θx\sin^3\theta+y\cos^3\theta=\sin\theta\cos\theta and xsin⁡θ=ycos⁡θx\sin\theta=y\cos\theta , then prove that x2+y2=1x^2+y^2=1 .

Show steps (8)

  1. Given xsin⁡θ=ycos⁡θx\sin\theta=y\cos\theta, so y=xsin⁡θcos⁡θy=\dfrac{x\sin\theta}{\cos\theta}.

  2. Substitute in xsin⁡3θ+ycos⁡3θ=sin⁡θcos⁡θx\sin^3\theta+y\cos^3\theta=\sin\theta\cos\theta:

  3. xsin⁡3θ+xsin⁡θcos⁡θ⋅cos⁡3θ=sin⁡θcos⁡θx\sin^3\theta+\frac{x\sin\theta}{\cos\theta}\cdot\cos^3\theta=\sin\theta\cos\theta

  4. xsin⁡3θ+xsin⁡θcos⁡2θ=sin⁡θcos⁡θx\sin^3\theta+x\sin\theta\cos^2\theta=\sin\theta\cos\theta

  5. xsin⁡θ(sin⁡2θ+cos⁡2θ)=sin⁡θcos⁡θx\sin\theta(\sin^2\theta+\cos^2\theta)=\sin\theta\cos\theta

  6. xsin⁡θ=sin⁡θcos⁡θ  ⇒  x=cos⁡θx\sin\theta=\sin\theta\cos\theta\;\Rightarrow\;x=\cos\theta

  7. Then y=xsin⁡θcos⁡θ=cos⁡θsin⁡θcos⁡θ=sin⁡θy=\dfrac{x\sin\theta}{\cos\theta}=\dfrac{\cos\theta\sin\theta}{\cos\theta}=\sin\theta.

  8. x2+y2=cos⁡2θ+sin⁡2θ=1x^2+y^2=\cos^2\theta+\sin^2\theta=1

Answer

x=cos⁡θ,  y=sin⁡θx=\cos\theta,\;y=\sin\theta and so x2+y2=1x^2+y^2=1, hence proved.

Q4. If acos⁡θ−bsin⁡θ=ca\cos\theta-b\sin\theta=c , then prove that (asin⁡θ+bcos⁡θ)=±a2+b2−c2(a\sin\theta+b\cos\theta)=\pm\sqrt{a^2+b^2-c^2} .

Show steps (8)

  1. Expand the sum of the squares of the two expressions.

  2. (acos⁡θ−bsin⁡θ)2+(asin⁡θ+bcos⁡θ)2(a\cos\theta-b\sin\theta)^2+(a\sin\theta+b\cos\theta)^2

  3. =a2cos⁡2θ−2absin⁡θcos⁡θ+b2sin⁡2θ+a2sin⁡2θ+2absin⁡θcos⁡θ+b2cos⁡2θ=a^2\cos^2\theta-2ab\sin\theta\cos\theta+b^2\sin^2\theta+a^2\sin^2\theta+2ab\sin\theta\cos\theta+b^2\cos^2\theta

  4. =a2(cos⁡2θ+sin⁡2θ)+b2(sin⁡2θ+cos⁡2θ)=a2+b2=a^2(\cos^2\theta+\sin^2\theta)+b^2(\sin^2\theta+\cos^2\theta)=a^2+b^2

  5. Given acos⁡θ−bsin⁡θ=ca\cos\theta-b\sin\theta=c, so

  6. c2+(asin⁡θ+bcos⁡θ)2=a2+b2c^2+(a\sin\theta+b\cos\theta)^2=a^2+b^2

  7. (asin⁡θ+bcos⁡θ)2=a2+b2−c2(a\sin\theta+b\cos\theta)^2=a^2+b^2-c^2

  8. Taking square root, asin⁡θ+bcos⁡θ=±a2+b2−c2a\sin\theta+b\cos\theta=\pm\sqrt{a^2+b^2-c^2}.

Answer

asin⁡θ+bcos⁡θ=±a2+b2−c2a\sin\theta+b\cos\theta=\pm\sqrt{a^2+b^2-c^2}, hence proved.

Q5. A bird is sitting on the top of a 80 m high …

A bird is sitting on the top of a 80 m high tree. From a point on the ground, the angle of elevation of the bird is 45∘45^\circ. The bird flies away horizontally in such away that it remained at a constant height from the ground. After 2 seconds, the angle of elevation of the bird from the same point is 30∘30^\circ. Determine the speed at which the bird flies. (3=1.732)(\sqrt{3}=1.732)

Show steps (7)

  1. Let PP be the point on the ground, bird first at BB (top of tree, height 8080 m), later at CC (same height 8080 m).

  2. In the first right triangle, tan⁡45∘=80PA\tan 45^\circ=\dfrac{80}{PA}.

  3. 1=80PA  ⇒  PA=80 m1=\frac{80}{PA}\;\Rightarrow\;PA=80\text{ m}

  4. In the second right triangle, tan⁡30∘=80PD\tan 30^\circ=\dfrac{80}{PD}.

  5. 13=80PD  ⇒  PD=803 m\frac{1}{\sqrt3}=\frac{80}{PD}\;\Rightarrow\;PD=80\sqrt3\text{ m}

  6. Distance flown =PD−PA=803−80=80(1.732−1)=80(0.732)=58.56=PD-PA=80\sqrt3-80=80(1.732-1)=80(0.732)=58.56 m.

  7. Speed=distancetime=58.562=29.28 m/s\text{Speed}=\frac{\text{distance}}{\text{time}}=\frac{58.56}{2}=29.28\text{ m/s}

Answer

Speed of the bird =29.28=29.28 m/s

Q6. An aeroplane is flying parallel to the Earth's surface at a speed …

An aeroplane is flying parallel to the Earth's surface at a speed of 175 m/sec and at a height of 600 m. The angle of elevation of the aeroplane from a point on the Earth's surface is 37∘37^\circ. After what period of time does the angle of elevation increase to 53∘53^\circ? (tan⁡53∘=1.3270, tan⁡37∘=0.7536)(\tan 53^\circ=1.3270,\ \tan 37^\circ=0.7536)

Show steps (7)

  1. Height of the plane =600=600 m (constant).

  2. At elevation 37∘37^\circ, horizontal distance d1d_1: tan⁡37∘=600d1\tan 37^\circ=\dfrac{600}{d_1}.

  3. d1=6000.7536=796.18 md_1=\frac{600}{0.7536}=796.18\text{ m}

  4. At elevation 53∘53^\circ, horizontal distance d2d_2: tan⁡53∘=600d2\tan 53^\circ=\dfrac{600}{d_2}.

  5. d2=6001.3270=452.15 md_2=\frac{600}{1.3270}=452.15\text{ m}

  6. Distance flown =d1−d2=796.18−452.15=344.03=d_1-d_2=796.18-452.15=344.03 m.

  7. Time=344.03175=1.97 seconds\text{Time}=\frac{344.03}{175}=1.97\text{ seconds}

Answer

About 1.971.97 seconds (nearly 22 seconds)

Q7. A bird is flying from AA towards BB at an angle of …

A bird is flying from AA towards BB at an angle of 35∘35^\circ, a point 30 km away from AA. At BB it changes its course of flight and heads towards CC on a bearing of 48∘48^\circ and distance 32 km away.

(i)

How far is BB to the North of AA? (sin⁡55∘=0.8192, cos⁡55∘=0.5736, sin⁡42∘=0.6691, cos⁡42∘=0.7431)(\sin 55^\circ=0.8192,\ \cos 55^\circ=0.5736,\ \sin 42^\circ=0.6691,\ \cos 42^\circ=0.7431)

  1. The bird flies from AA to BB, AB=30AB=30 km, making 35∘35^\circ with the north direction (towards the west).

  2. Northward distance =ABcos⁡35∘=30sin⁡55∘=AB\cos 35^\circ=30\sin 55^\circ.

  3. =30×0.8192=24.576 km=30\times 0.8192=24.576\text{ km}

Answer

BB is 24.57624.576 km north of AA

(ii)

How far is BB to the West of AA? (sin⁡55∘=0.8192, cos⁡55∘=0.5736, sin⁡42∘=0.6691, cos⁡42∘=0.7431)(\sin 55^\circ=0.8192,\ \cos 55^\circ=0.5736,\ \sin 42^\circ=0.6691,\ \cos 42^\circ=0.7431)

  1. Westward distance =ABsin⁡35∘=30cos⁡55∘=AB\sin 35^\circ=30\cos 55^\circ.

  2. =30×0.5736=17.208 km=30\times 0.5736=17.208\text{ km}

Answer

BB is 17.20817.208 km west of AA

(iii)

How far is CC to the North of BB? (sin⁡55∘=0.8192, cos⁡55∘=0.5736, sin⁡42∘=0.6691, cos⁡42∘=0.7431)(\sin 55^\circ=0.8192,\ \cos 55^\circ=0.5736,\ \sin 42^\circ=0.6691,\ \cos 42^\circ=0.7431)

  1. From BB the bearing is 48∘48^\circ, i.e. BC=32BC=32 km makes 48∘48^\circ with the north direction.

  2. Northward distance =BCcos⁡48∘=32sin⁡42∘=BC\cos 48^\circ=32\sin 42^\circ.

  3. =32×0.6691=21.4112 km=32\times 0.6691=21.4112\text{ km}

Answer

CC is 21.4121.41 km north of BB

(iv)

How far is CC to the East of BB? (sin⁡55∘=0.8192, cos⁡55∘=0.5736, sin⁡42∘=0.6691, cos⁡42∘=0.7431)(\sin 55^\circ=0.8192,\ \cos 55^\circ=0.5736,\ \sin 42^\circ=0.6691,\ \cos 42^\circ=0.7431)

  1. Eastward distance =BCsin⁡48∘=32cos⁡42∘=BC\sin 48^\circ=32\cos 42^\circ.

  2. =32×0.7431=23.7792 km=32\times 0.7431=23.7792\text{ km}

Answer

CC is 23.7823.78 km east of BB

Q8. Two ships are sailing in the sea on either side of the …

Two ships are sailing in the sea on either side of the lighthouse. The angles of depression of two ships as observed from the top of the lighthouse are 60∘60^\circ and 45∘45^\circ respectively. If the distance between the ships is 200(3+13)200\left(\dfrac{\sqrt{3}+1}{\sqrt{3}}\right) metres, find the height of the lighthouse.

Show steps (8)

  1. Let AB=hAB=h be the height of the lighthouse, with ships at CC and DD on either side.

  2. Angle of depression 60∘60^\circ gives elevation 60∘60^\circ at CC: tan⁡60∘=hBC\tan 60^\circ=\dfrac{h}{BC}.

  3. BC=h3BC=\frac{h}{\sqrt3}

  4. Angle of depression 45∘45^\circ gives tan⁡45∘=hBD\tan 45^\circ=\dfrac{h}{BD}.

  5. BD=hBD=h

  6. Distance between the ships:

  7. BC+BD=h3+h=h(3+13)BC+BD=\frac{h}{\sqrt3}+h=h\left(\frac{\sqrt3+1}{\sqrt3}\right)

  8. Given this equals 200(3+13)200\left(\dfrac{\sqrt3+1}{\sqrt3}\right), so h=200h=200.

Answer

Height of the lighthouse =200=200 m

Q9. A building and a statue are in opposite side of a street …

A building and a statue are in opposite side of a street from each other 35 m apart. From a point on the roof of building the angle of elevation of the top of statue is 24∘24^\circ and the angle of depression of base of the statue is 34∘34^\circ. Find the height of the statue. (tan⁡24∘=0.4452, tan⁡34∘=0.6745)(\tan 24^\circ=0.4452,\ \tan 34^\circ=0.6745)

Show steps (7)

  1. Let PP be the point on the roof of the building, CDCD the statue, width of street =35=35 m.

  2. Draw the horizontal from PP meeting the statue at EE; then PE=35PE=35 m.

  3. Top part of statue: tan⁡24∘=CE35\tan 24^\circ=\dfrac{CE}{35}.

  4. CE=35×0.4452=15.582 mCE=35\times 0.4452=15.582\text{ m}

  5. Lower part (depression of base): tan⁡34∘=ED35\tan 34^\circ=\dfrac{ED}{35}.

  6. ED=35×0.6745=23.6075 mED=35\times 0.6745=23.6075\text{ m}

  7. Height of statue =CE+ED=15.582+23.6075=39.1895=CE+ED=15.582+23.6075=39.1895 m.

Answer

Height of the statue =39.19=39.19 m (approximately)

Quick check

  1. 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 1 of 5

  2. 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}
    Show answer

    Answer: (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.

    Question 2 of 5

  3. (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 3 of 5

  4. 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 4 of 5

  5. 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}}
    Show answer

    Answer: (2) b3\dfrac{b}{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}.

    Question 5 of 5
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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