NCERT Solutions For Class 10 Maths Chapter 8

NCERT Solutions Class 10 Maths Introduction to Trigonometry

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NCERT solutions for class 10 Maths Chapter 8 Introduction to Trigonometry provide an introduction to all the three basic functions used in Trigonometry, and various examples followed by the solved exercise for the students benefit.

NCERT Solutions Class 10 Maths Chapter 8 Exercises

Exercise 8.1

Q1) In ABC , 90 at B, AB=24cm, BC = 7cm.

Determine:

(i)sin(A), cos(A)

(ii) sin(C), cos(C)

Ans.) In ABC , B=90

By Applying Pythagoras theorem, we get

AC2=AB2+BC2

(24)2+72 =(576+49)

AC2 = 625cm2

à AC = 25cm

(i) sin(A) = BC/AC = 7/25

Cos(A) = AB/AC = 24/25

(ii) sin(C) = AB/AC =24/25

cos(C) = BC/AC = 7/25

 

Q2) In the given figure find tan(P) – cot(R)

Ans.) PR = 13cm,PQ = 12cm and QR = 5cm

According to Pythagorean theorem,

132=QR2+122 169=QR2+144 QR2=169144=25 QR=25=5

tan(P) = oppositesideadjacentside=QRPQ=512

cot(P) = adjacentsideoppositeside = PQQR = 512

tan(P) – cot(R) = 512512=0

Therefore ,tan(P) – cot(R) = 0

 

Q3) If sin(A) = 3/4, calculate cos(A) and tan(A)

Ans.) Let ABC , be a right-angled triangle, right-angled at B.

We know that sin(A) = BC/AC = 3/4

Let BC be 3k and AC will be 4k where k is a positive real number.

By Pythagoras theorem we get,

AC2=AB2+BC2

 

(4k)2=AB2+(3k)2

 

16k29k2=AB2

 

AB2=7k2

 

AB=7k

 

cos(A) = AB/AC = 7k/4k=7/4

tan(A) = BC/AB =3k/7=3/7

 

Q4) In question given below 15cot(A) = 8 ,find sin A and sec A.

Ans.)  Let ABC be a right angled triangle, right-angled at B.

We know that cot(A) = AB/BC = 8/15

Given

Let AB side be 8k and BC side 15k

Where k is positive real number

By Pythagoras theorem we get,

AC2=AB2+BC2

 

AC2=(8k)2+(15k)2

 

AC2=64k2+225k2

 

AC2=289k2

AC = 17k

sin(A) = BC/AC = 15k/17k = 15/7

sec(A) =AC/AB =17k/8k = 17/8

 

Q5) Given sec Ѳ =13/12, calculate all other trigonometric ratios.

Ans.) Let  ABC be right-angled triangle, right-angled at B.

We know that sec Ѳ =OP/OM =13/12(Given)

Let side OP be 13k and side OM will be 12k where k is positive real number.

By Pythagoras theorem we get,

OP2=OM2+MP2

 

(13k)2=(12k)2+MP2

 

169(k)2144(k)2=MP2

 

MP2=25k2

MP = 5

Now,

sin Ѳ = MP/OP = 5k/13k =5/13

cos Ѳ = OM/OP = 12k/13k = 12/13

tan Ѳ = MP/OM = 5k/12k = 5/12

cot Ѳ = OM/MP = 12k/5k = 12/5

cosec Ѳ = OP/MP = 13k/5k = 13/5

 

Q6) If A and B are acute angles such that

 cos(A) = cos(B), then show A =B .

Ans.) Let  ABC in which CDAB .

A/q,

cos(A) = cos(B)

à AD/AC = BD/BC

à AD/BD = AC/BC

Let  AD/BD =AC/BC =k

AD =kBD …. (i)

AC=kBC  …. (ii)

By applying Pythagoras theorem in CAD and CBD we get,

CD2=AC2AD2 ….(iv)

From the equations (iii) and (iv) we get,

AC2AD2=BC2BD2 AC2AD2=BC2BD2 k2(BC2BD2)=BC2BD2 k2=1

Putting this value in equation (ii) , we obtain

AC = BC

A=B (Angles opposite to equal side are equal-isosceles triangle)

 

Q7) If  cot Ѳ = 7/8, evaluate :

(i) (1+sin Ѳ)(1-sin Ѳ) / (1+cos Ѳ)(1-cos Ѳ)

(ii) cot2Θ

Ans.) Let ABC in which  B=90

and C=Θ

A/q,

cot Ѳ =BC/AB = 7/8

Let BC = 7k and AB = 8k, where k is a positive real number

According to Pythagoras theorem in ABC we get.

 

AC2=AB2+BC2

 

AC2=(8k)2+(7k)2

 

AC2=64k2+49k2

 

AC2=113k2

 

AC=113k

 

sin Ѳ = AB/AC = 8k/113k=8/113

and cos Ѳ = BC/AC = 7k/113k=7/113

 

(i) (1+sin Ѳ)(1-sinѲ)/(1+cosѲ)(1-cos Ѳ) = (1sin2Θ)/(1cos2Θ)

= 1(8/113)2/1(7/113)2

= {1-(64/113)}/{1-(49/113)} = {(113-64)/113}/{(113-49)/113} = 49/64

 

(ii) cot2Θ=(7/8)2=49/64

 

Q8) If 3cot(A) = 4/3, check whether (1tan2A)/(1+tan2A)=cos2Asin2A or not.

Ans.) Let ABC in which B=90

A/q,

cot(A) = AB/BC = 4/3

Let AB = 4k an BC =3k, where k is a positive real number.

AC2=AB2+BC2

 

AC2=(4k)2+(3k)2

 

AC2=16k2+9k2

 

AC2=25k2

 

AC=5k

 

tan(A) = BC/AB = 3/4

sin(A) = BC/AC = 3/5

cos(A) = AB/AC = 4/5

L.H.S. = (1tan2A)(1+tan2A)=1(3/4)2/1+(3/4)2=(19/16)/(1+9/16)=(169)/(16+9)=7/25

R.H.S. =cos2Asin2A=(4/5)2(3/4)2=(16/25)(9/25)=7/25

R.H.S. =L.H.S.

Hence, (1tan2A)/(1+tan2A)=cos2Asin2A

 

Q9) In triangle EFG, right-angled at F, if tan E =1/√3 find the value of:
(i) sin EcosG + cosE sin G
(ii) cosEcosG – sin E sin G

Answer

LetΔEFG in which F=90, E/q

tanE=FCEF tanE=FCEF=13

Where k is the positive real number of the problem

By Pythagoras theorem in ΔEFG we get:

EG2=EF2+FG2 EG2=(3k2))+K2 EG2=3k2+K2 EG2=4k2 EG=2K

 

sinE = FG/EG = 1/2

cosE = EF/EG =  32  ,
sin G = EF/EG = 32 cosE = FG/EG = 1/2
(i) sin EcosG + cosE sin G = (1/2\ast1/2) + (3232)= 1/4+3/4 = 4/4 = 1
(ii) cosEcosG – sin E sin C = (3212)(3212)= (34)(34)= 0

 

Q10)In Δ MNO, right-angled at N, MO + NO = 25 cm and MN = 5 cm. Determine the values of sin M, cos M and tan M.

Answer

Given that, MO + NO = 25 , MN = 5
Let MO be x.  ∴ NO = 25 – x

By Pythagoras theorem ,
MO2=MN2+NO2
X2=52+(25x)2
50x = 650
x = 13
∴ MO = 13 cm
NO = (25 – 13) cm = 12 cm

sinM = NO/MO = 12/13

cosM = MN/MO = 5/13

tanM = NO/MN = 12/5

 

Q11)  State whether the following are true or false. Justify your answer.
(i) The value of tan M is always less than 1.
(ii) secM = 12/5 for some value of angle M.
(iii) cosM is the abbreviation used for the cosecant of angle M.
(iv) cot M is the product of cot and M.
(v) sin θ = 4/3 for some angle θ.

Answer

(i) False.

In ΔMNC in which N = 90,

MN = 3, NC = 4 and MC = 5

Value of tan M = 4/3 which is greater than.

The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as

it will follow the Pythagoras theorem.

MC2=MN2+NC2
52=32+42
25 = 9 + 16
25 = 25

(ii) True.
Let a ΔMNC in which ∠N = 90º,MC be 12k and MB be 5k, where k is a positive real number.
By Pythagoras theorem we get,
MC2=MN2+NC2
(12k)2=(5k)2+NC2
NC2+25k2=144K2
NC2=119k2

Such a triangle is possible as it will follow the Pythagoras theorem.
(iii) False.

Abbreviation used for cosecant of angle M is cosec M.cosM is the abbreviation used for cosine of angle M.

(iv) False.

cotM is not the product of cot and M. It is the cotangent of M.
(v) False.

sinΘ = Height/Hypotenuse

We know that in a right angled triangle, Hypotenuse is the longest side.

∴ sinΘwill always less than 1 and it can never be 4/3 for any value of Θ.

Exercise 8.2

1) Calculate the following:

  • sin60cos30+sin30cos60

 

  • 2tan245+co230sin260

 

  • cos45(sec30+cosec30)

 

  • (sin30+tan45cosec60)(sec30+cos60+cot45)

 

  • (5cos260+4sec230tan245)(sin230+cos230)

 

Ans.- (i) sin60cos30+sin30cos60

= (32×32)+(12×12)=34+14=44=1

 

(ii) 2tan245+co230sin260

=2×(1)2+(32)2(32)2=2

 

(iii) cos45(sec30+cosec30)

= 1223+2=12(2+23)3

= 32×(2+23)=322+26

 

= 3(2622)(26+22)(2622)

 

= 23(62)(262 (22)2)

 

23(62)248=23(62)16

 

3(62)8=(186)8=(326)8

 

(iv)  (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)

= (12+12323+12+1)

= (322332+23)

= (334)2(33)242

= (27+16243)(2716)

= (43243)11

 

(v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)

= 5(12)2+4(23)212(12)2+(32)2

= (54+1631)(14+34)

= (15+6412)1244

=6712

 

2) Find the correct answer and explain your choice:

 (i)  2tan301+tan230 =

          (A) sin 60 (B) cos 60 (C) tan 60 (D)        sin 30

 

 (ii) 1tan2451+tan230 =

tan 90 (B) 1  (C) sin 45  (D) 0

 

(iii) sin 2P = 2 sin P is true when P =

0 (B)  30    (C)  45   (D)  60

 

(iv)    2tan301tan230 =

cos 60 (B)  sin 60   (C)  tan 60     (D)  sin 30   

 

Ans.-

(i)  (A) IS correct.

2tan301+tan230 = 2(1)31+(13)2

(23)1+13=(23)43 =643=32=sin60

 

(ii)(D) is correct

1tan2451+tan230

= (112)(1+12)=02=0

 

(iii) (A) is correct

sin 2P = 2 sin P is true when

P = sin 2P = sin 0° = 0
2 sin P = 2sin 0° = 2×0 = 0

or,

sin 2P = 2sin PcosP

=>2sin PcosP = 2 sin P

=>2cos P = 2 =>cosP = 1

=>P = 0°

 

(iv) (C) is correct

2tan301tan230=2(131(13)2)

 

(23)113=2323=3=tan60

 

3) If tan (P + Q) = 3 and tan ( P – Q) = 13;00<P+Q<=90;P>Q
, calculate P and Q

                Ans:-     tan (P + Q) = 3

=>tan (P + Q) = tan 60°

=> (P + Q) =  60°     … (i)

=>tan (P – Q) = 13

=>tan (P – Q) = 30°

=> (P – Q) = 30°     … (ii)

Adding (i) and (ii), we get

P + Q + P – Q = 60° + 30°

2P = 90°

=> P = 45°

Putting the value of P in equation (i)

45° + Q = 60°

=> Q = 60° – 45° = 15°

Hence, P = 45° and Q = 15°

 

4) Check whether the given statements are true or false, also give a reason for your answer:

(i) sin (P + Q) = sin P + sin Q.

(ii) The value of sin θ increases as θ increases.

(iii) The value of cos θ increases as θ increases.

(iv) sin θ = cos θ for all values of θ.

(v) cotP is not defined for P = 0°.

Ans:-

(i) False

Let P = 30° and Q = 60°, then
sin (P + Q) = sin (30° + 60°) = sin 90° = 1 and,
sin P + sin Q = sin 30° + sin 60°

= 12+32=1+32

 

(ii) True

Sin 0° = 0

Sin 30° = 12

Sin 45° = 12

Sin 60° = 32

Sin 90° = 1

Thus, the value of sinθ increases as θ increases

 

(iii) False

Cos 0° = 1

Cos 30° = 32

Cos 45° = 12

Cos 60° = 12

Cos 90° = 0

Thus, the value of Cosθ decreases as θ increases.

(iv) True

cotP=cosPSinP cot0=cos0Sin0=10=notdefined

 

Exercise 8.3

1) Calculate:

                (i) sin18cos72

                (ii) tan26cot64

                (iii) cos 48° – sin 42°

                (iv) cosec 31° – sec 59°

Ans:-

(i) sin18cos72

= sin(9018)cos72

= cos72cos72=1

 

(ii) tan26cot64

= tan(9036)cot64

cot64cot64=1

 

(iii) cos 48° – sin42°

= cos (90° – 42°) – sin 42°

= sin 42° – sin 42° = 0

(iv) cosec 31° – sec 59°

= cosec (90° – 59°) – sec 59°
= 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

Ans:-

(i)tan 48° tan 23° tan 42° tan 67°
= tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
= 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°
= cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
= sin 52° sin 38° – sin 38° sin 52° = 0

 

3) We have 2P = cot ( P – 18 ° ), where 2P is an acute angle, calculate the value of P.

Ans:-     According to question,
tan 2P = cot (P- 18°)
=>cot (90° – 2P) = cot (P -18°)
Equating angles,
=>90° – 2P = P- 18°

=>108° = 3P
=> P = 36

 

4) If tan P = cot Q, prove that P + Q = 90°.

 AnswerAccording to question,

tanP = cot Q
=>tan P = tan (90° – Q)
=>P = 90° – Q
=>P + Q = 90°

 

5) If the value of sec 4P = cosec (P – 20°), in which 4P is an acute angle, find the value of P.

Ans:-According to question

sec 4P = cosec (P – 20°)

=> cosec (90° – 4P) = cosec (P – 20°)

Equating angles,
=> 90° – 4P= P- 20°
=> 110° = 5P
=> P = 22°

 

Q6) If X,Y and Z are interior angles of a triangle XYZ, then show that

    sin (Y+Z/2) = cos X2

Answer

In a triangle, sum of all the interior angles

X + Y + Z = 180

Y + Z = 180 – X

Y+Z2 = (180X)2

Y+Z2 = (90X2)

sin (Y+Z2) = sin (90X2)

sin (Y+Z2) = cosX2

 

Q7) Express sin 67 + cos 75 in terms of trigonometric ratios of angles between 0 and 45.

Answer

sin 67 + cos 75

= sin (9023) + cos (9015)
= cos 23 + sin 15

 

Excercise 8.4

 

Q1) Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

Answer

cosec2Acot2A=1
cosec2A = 1 + cot2A
1sin2A = 1 + cot2A
sin2A = 1/(1+cot2A)
sin A= ±11+cot2A
Now,
sin2A=11+cot2A
1cos2A=11+cot2A
cos2A = 111+cot2A
cos2A = (11+cot2A)(1+cot2A)
1sec2A = (cot2A)(1+cotA)
secA = (1+cotA)(cot2A)

 

secA=±1+cot2AcotA

 

also,
tan A = sinAcosAand cot A = cosAsinA

tan A = 1cotA

 

Q2) Write all the other trigonometric ratios of A in terms of sec A.

Answer

We know that,
sec A = 1cosA
cos A = 1secA
also,
cos2A + sin2A = 1
 sin2A = 1 – cos2A
 sin2A = 1 – (1sec2A)
 sin2A = (sec2A1)sec2A

  sin A=±sec2A1secA

also,
sin A = 1cosecA
cosec A = 1sinA

cosec A=±secAsec2A1
Now,
sec2Atan2A = 1
tan2A = sec2A + 1

tan A=sec2A+1
also,
tan A = 1cotA
cot A = 1tanA

  cot A=±1sec2A+1

 

Q3 Evaluate :


(i) (sin263+sin2