RS Aggarwal Solutions for Class 10 Maths Chapter 10–Trignometric Ratios
RS Aggarwal Solutions for Class 10 Maths Chapter 10–Trignometric Ratios

Class 10: Maths Chapter 10 solutions. Complete Class 10 Maths Chapter 10 Notes.

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RS Aggarwal Solutions for Class 10 Maths Chapter 10–Trignometric Ratios

RS Aggarwal 10th Maths Chapter 1, Class 10 Maths Chapter 10 solutions

Page No 546:

Question 1:

If sin θ=32, find the value of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that sin θ = perpendicularhypotenuse= ABAC = 32 .

So, if AB = 3k, then AC = 2k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 
⇒ BC2 = AC2 AB2 = (2k)2  (3k)2
⇒ BC2 = 4k2 3k2 = k2
⇒ BC = k
Now, finding the other T-ratios using their definitions, we get:
   cos θ  = BCAC = k2k = 12
   tan θ  = ABBC = 3kk = 3

 ∴ cot θ  = 1tan θ = 13, cosec θ = 1sin θ = 23 and sec θ  = 1cos θ = 2

Page No 546:

Question 2:

If cos θ=725  find the values of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ .
Now, we know that cos θ = Basehypotenuse = BCAC  = 725 .

So, if BC = 7k, then AC = 25k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 
⇒ AB2 = AC2 BC2 = (25k)2  (7k)2.
⇒ AB2 = 625k2 49k2 = 576k2
⇒ AB = 24k
Now, finding the other trigonometric ratios using their definitions, we get:
   sin θ = ABAC  = 24k25k = 2425 
   tan θ = ABBC = 24k7k = 247 
 ∴ cot θ = 1tan θ = 724 , cosec θ = 1sin θ = 2524  and sec θ  = 1cos θ = 257 

Page No 546:

Question 3:

If tan θ=158 find the values of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that tan θ = PerpendicularBase = ABBC = 158.

So, if BC = 8k, then AB = 15k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (15k)2 + (8k)2
⇒ AC2 = 225k2 + 64k2 = 289k2
⇒ AC = 17k

Now, finding the other T-ratios using their definitions, we get:
  sin θ  = ABAC = 15k17k = 1517
  cos θ  = BCAC = 8k17k = 817

∴ cot θ  = 1tan θ = 815, cosec θ = 1sin θ = 1715 and sec θ  = 1cos θ = 178

Page No 546:

Question 4:

If cot θ = 2, find the value of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that cot θbasePerpendicular = BCAB = 2.


So, if BC = 2k, then AB = k, where k is a positive number.
Now, using Pythagoras theorem, we have:
AC2 = AB2 + BC2 = (2k)2 + (k)2
⇒ AC2 = 4k2 + k2 = 5k2
⇒ AC = 5k
Now, finding the other T-ratios using their definitions, we get:
   sin θ  = ABAC = k5k = 15
   cos θ  = BCAC = 2k5k = 25

∴ tan θ  = 1cot θ = 12, cosec θ = 1sin θ = 5 and sec θ  = 1cos θ = 52

Page No 546:

Question 5:

If cosec θ = 10, the find the values of all T-ratios of θ.

Answer:

Let us first draw a right ABC, right angled at B and C=θ.
Now, we know that cosec θ = HypotenusePerpendicular = ACAB= 101.

So, if AC = (10)k, then AB = k, where k is a positive number.
Now, by using Pythagoras theorem, we have:
AC2 = AB2 + BC2 
⇒ BC2 = AC2 AB2 = 10k2 k2
⇒ BC2 = 9k2
⇒ BC = 3k
Now, finding the other T-ratios using their definitions, we get:
   tan θ  = ABBC = k3k = 13

   cos θ  = BCAC = 3k10k = 310

 ∴ sin θ=1cosec θ=110, cot θ  = 1tan θ = 3 and sec θ  = 1cos θ = 103

Page No 546:

Question 6:

If sinθ=a2b2a2+b2, find the values of all T-ratios of θ.

Answer:

We have sinθ=a2b2a2+b2,

As,

cos2θ=1sin2θ=1a2b2a2+b22=11a2b22a2+b22=a2+b22a2b22a2+b22=a2+b2a2b2a2+b2+a2b2a2+b22
=a2+b2a2+b2a2+b2+a2b2a2+b22=2b22a2a2+b22cos2θ=4a2b2a2+b22cosθ=4a2b2a2+b22cosθ=2aba2+b2

Also,

tanθ=sinθcosθ=a2b2a2+b22aba2+b2=a2b22ab

Now,

cosecθ=1sinθ=1a2b2a2+b2=a2+b2a2b2

Also,

secθ=1cosθ=12aba2+b2=a2+b22ab

And,

cotθ=1tanθ=1a2b22ab=2aba2b2

Page No 546:

Question 7:

If sinθ=cc2+d2, where d > 0 then find the values of cos θ and tan θ.

Answer:

Given: sinθ=cc2+d2Since, sinθ=PHP=c and H=c2+d2Using Pythagoras theorem,P2+B2=H2c2+B2=c2+d2B2=d2B=dTherefore,cosθ=BH=dc2+d2tanθ=PB=cdHence, cosθ=dc2+d2 and tanθ=cd.

Page No 546:

Question 8:

If 3 tan θ=1 then evaluate (cos2θ – sin2θ).

Answer:

Given: 3tanθ=1tanθ=13Since, tanθ=PBP=1 and B=3Using Pythagoras theorem,P2+B2=H212+32=H2H2=1+3=4H=2Therefore,sinθ=PH=12cosθ=BH=32cos2θsin2θ=322122                   =3414=24                   =12Hence, cos2θsin2θ=12.

Page No 546:

Question 9:

If 4tan θ = 3 then prove that sin θ cos θ=1225.

Answer:

Given: 4tanθ=3tanθ=34Since, tanθ=PBP=3 and B=4Using Pythagoras theorem,P2+B2=H232+42=H2H2=9+16=25H=5Therefore,sinθ=PH=35cosθ=BH=45sinθ×cosθ=35×45                   =1225Hence, sinθ×cosθ=1225.

Page No 546:

Question 10:

If sinθ=ab, show that secθ+tanθ=b+aba.

Answer:

LHS=secθ+tanθ=1cosθ+sinθcosθ=1+sinθcosθ=1+sinθ1sin2θ=1+ab1ab2
=11+ab11a2b2=b+abb2a2b2=b+abb2a2b=b+ab+aba
=b+ab+aba=b+aba=b+aba=RHS

Page No 547:

Question 11:

If tan θ = abab, show that (a sinθb cosθa sinθ+b cosθ)=(a2b2)(a2+b2).a sinθb cosθa sinθ+b cosθ=a2b2a2+b2.

Answer:

It is given that tan θ = abθ = ab.

LHS = a sinθ  b cosθa sinθ + b cosθa sinθ  b cosθa sinθ + b cosθ
 Dividing the numerator and denominator by cos θθ, we get:

 a tan θ  ba tan θ + ba tan θ  ba tan θ + b       (∵ tan θ = sin θcos θθ = sin θcos θ)
Now, substituting the value of tan θθ in the above expression, we get:
 a(ab)  ba(ab) + b= a2b  ba2b + b= a2  b2a2 + b2 = RHSaab  baab + b= a2b  ba2b + b= a2  b2a2 + b2 = RHS
  i.e., LHS = RHS

 Hence proved.

Page No 547:

Question 12:

If sin θ=1213sin θ=1213 then evaluate (2sin θ−3cos θ4sin θ−9cos θ)2sin θ3cos θ4sin θ9cos θ.

Answer:

Given: sinθ=1213Since, sinθ=PHP=12 and H=13Using Pythagoras theorem,P2+B2=H2122+B2=132B2=169144B2=25B=5Therefore,cosθ=BH=513Now,(2sinθ3cosθ4sinθ9cosθ)=(2(1213)3(513)4(1213)9(513))                        =(2413151348134513)                        =(24154845)                        =93                        =3Hence, (2sinθ3cosθ4sinθ9cosθ)=3.Given: sinθ=1213Since, sinθ=PHP=12 and H=13Using Pythagoras theorem,P2+B2=H2122+B2=132B2=169144B2=25B=5Therefore,cosθ=BH=513Now,2sinθ3cosθ4sinθ9cosθ=212133513412139513                        =2413151348134513                        =24154845                        =93                        =3Hence, 2sinθ3cosθ4sinθ9cosθ=3.

Page No 547:

Question 13:

If tan θ=12tan θ=12 then evaluate (cos θsin θ+sin θ1+cos θ)cos θsin θ+sin θ1+cos θ

Answer:

Given: tanθ=12Since, tanθ=PBP=1 and B=2Using Pythagoras theorem,P2+B2=H212+22=H2H2=1+4H2=5H=5Therefore,sinθ=PH=15cosθ=BH=25Now,(cosθsinθ+sinθ1+cosθ)=(2515+151+25)                             =21+155+25                             =(21+15+2)                             =(2+(15+2×5252))                             =(2+(5254))                             =(2+52)                             =5Hence, (cosθsinθ+sinθ1+cosθ)=5.Given: tanθ=12Since, tanθ=PBP=1 and B=2Using Pythagoras theorem,P2+B2=H212+22=H2H2=1+4H2=5H=5Therefore,sinθ=PH=15cosθ=BH=25Now,cosθsinθ+sinθ1+cosθ=2515+151+25                             =21+155+25                             =21+15+2                             =2+15+2×5252                             =2+5254                             =2+52                             =5Hence, cosθsinθ+sinθ1+cosθ=5.

Page No 547:

Question 14:

If sinα=12sinα=12, prove that (3cosα−4cos3α)=03cosα4cos3α=0.

Answer:

LHS=(3cosα4cos3α)=cosα(34cos2α)=1sin2α[34(1sin2α)]=1(12)2[34(1(12)2)]=1114[34(1114)]=34[34(34)]=34[33]=34[0]=0=RHSLHS=3cosα4cos3α=cosα34cos2α=1sin2α341sin2α=1122341122=1114341114=343434=3433=340=0=RHS

Page No 547:

Question 15:

If 3 cot θ = 2, show that (4sinθ3cosθ2sinθ+6cosθ)=13.4sinθ3cosθ2sinθ+6cosθ=13.

Answer:

It is given that cot θ = 23θ = 23.

LHS  = 4 sinθ  3 cosθ2 sinθ + 6 cosθ4 sinθ  3 cosθ2 sinθ + 6 cosθ
Dividing the above expression by sin θθ, we get:
4  3 cot θ2 + 6 cot θ4  3 cot θ2 + 6 cot θ                     [∵ cot θ = cosθsinθθ = cosθsinθ]
Now, substituting the values of cot θθ in the above expression, we get:
 4  3(23)2 + 6(23)= 4  22 + 4 = 26=13 4  3232 + 623= 4  22 + 4 = 26=13
 i.e., LHS = RHS
 
Hence proved.

Page No 547:

Question 16:

If sec θ = 178178 then prove that 34sin2θ4cos2θ3=3tan2θ13tan2θ34sin2θ4cos2θ3=3tan2θ13tan2θ.

Answer:

It is given that sec θθ = 178178.

Let us consider a right ABC right angled at B and C=θC=θ.
We know that cos θθ = 1sec θ= 817 = BCAC1sec θ= 817 = BCAC
 
So, if BC = 8k, then AC = 17k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2 BC2 = (17k)2 (8k)2
⇒ AB2 = 289k2 64k2 = 225k2
⇒ AB = 15k.

Now, tan θθ  = ABBC = 158ABBC = 158 and sin θθ = ABAC = 15k17k= 1517ABAC = 15k17k= 1517

The given expression is 3  4sin2θ4cos2θ 3 = 3  tan2θ1  3tan2θ3  4sin2θ4cos2θ 3 = 3  tan2θ1  3tan2θ.
 
 Substituting the values in the above expression, we get:
 LHS= 3  4(1517)24(817)2  3 = 3  900289256289 3 = 867900256867= 33611=33611LHS= 3  41517248172  3 = 3  900289256289 3 = 867900256867= 33611=33611

RHS = 3(158)213(158)2=322564167564=19222564675=33611=33611RHS = 31582131582=322564167564=19222564675=33611=33611

∴ LHS = RHS
Hence proved.

Page No 547:

Question 17:

If tan θ = 20212021, show that(1sinθ+cosθ)(1+sinθ+cosθ)=37.1sinθ+cosθ1+sinθ+cosθ=37.

Answer:

Let us consider a right ABC right angled at B and C=θC=θ.
Now, we know that tan θθABBCABBC = 20212021

So, if AB = 20k, then BC = 21k, where k is a positive number.
Using Pythagoras theorem, we get:
 AC2 = AB2 + BC2
⇒ AC2= (20k)2 + (21k)2
⇒ AC2 = 841k2
⇒  AC = 29k
Now, sin θθ = ABAC = 2029ABAC = 2029 and cos θθ = BCAC = 2129BCAC = 2129

Substituting these values in the given expression, we get:
  LHS=1  sinθ + cosθ1 + sinθ + cosθ= 1  2029 + 21291 + 2029 + 2129= 29  20 + 212929 + 20 + 2129= 3070 = 37 = RHS LHS=1  sinθ + cosθ1 + sinθ + cosθ= 1  2029 + 21291 + 2029 + 2129= 29  20 + 212929 + 20 + 2129= 3070 = 37 = RHS
∴ LHS = RHS

Hence proved.

Page No 547:

Question 18:

If tan θ = 1717 then prove that (cosec2θ+sec2θcosec2θsec2θ)=43.cosec2θ+sec2θcosec2θsec2θ=43.

Answer:

Let us consider a right ABC, right-angled at B and C=θ.
Now it is given that tan θABBC17.

So, if AB = k, then BC = 7k, where k is a positive number.
Using Pythagoras theorem, we have:
AC2 = AB2 + BC2
⇒ AC2 = (k)2 + (7k)2
⇒ AC2 = k2 + 7k2
⇒ AC = 22k
Now, finding out the values of the other trigonometric ratios, we have:
sin θ  = ABAC = k22k = 122
cos θ  = BCAC = 7 k22k = 722
∴ cosec θ  = 1sin θ = 22 and sec θ   = 1cos θ = 227
Substituting the values of cosec θ  and sec θ  in the given expression, we get:
 cosec2θ – sec2θcosec2θ + sec2θ=(22)2 – 2272(22)2 + 2272=8 – 878 + 87=56 – 8756 + 87=4864 = 34 = RHS
 i.e., LHS = RHS
 
Hence proved.

Page No 547:

Question 19:

If sinθ=34sinθ=34, show that cosec2θcot2θsec2θ1=73cosec2θcot2θsec2θ1=73.

Answer:

LHS=cosec2θcot2θsec2θ1=1tan2θ=cot2θ=cotθ=cosec2θ1=(1sinθ)21=(1(34))21=(43)21=1691=1699=79=73=RHSLHS=cosec2θcot2θsec2θ1=1tan2θ=cot2θ=cotθ=cosec2θ1=1sinθ21=13421=4321=1691=1699=79=73=RHS

Page No 547:

Question 20:

If 3 tan A = 4 then prove that
(i) sec Acosec Asec A+cosec A=17sec Acosec Asec A+cosec A=17
(ii) 1sin A1+cos A=1221sin A1+cos A=122

Answer:

(i)
 ​LHS=secθcosecθsecθ+cosecθ=(1cosθ1sinθ)(1cosθ+1sinθ)=(sinθcosθsinθ cosθ)(sinθ+cosθsinθ cosθ)=(sinθcosθsinθ)(sinθ+cosθsinθ)=(sinθsinθcosθsinθ)(sinθsinθ+cosθsinθ)=1cotθ1+cotθ=(134)(1+34)LHS=secθcosecθsecθ+cosecθ=1cosθ1sinθ1cosθ+1sinθ=sinθcosθsinθ cosθsinθ+cosθsinθ cosθ=sinθcosθsinθsinθ+cosθsinθ=sinθsinθcosθsinθsinθsinθ+cosθsinθ=1cotθ1+cotθ=1341+34
=(14)(74)=17=17=RHS=1474=17=17=RHS

(ii)
Given: 3tanA=4tanA=43Since, tanA=PBP=4 and B=3Using Pythagoras theorem,P2+B2=H242+32=H2H2=16+9H2=25H=5Therefore,sinA=PH=45cosA=BH=35Now,1sinA1+cosA=1451+35                  =5455+35                  =1585                  =18                  =122Hence, 1sinA1+cosA=122.Given: 3tanA=4tanA=43Since, tanA=PBP=4 and B=3Using Pythagoras theorem,P2+B2=H242+32=H2H2=16+9H2=25H=5Therefore,sinA=PH=45cosA=BH=35Now,1sinA1+cosA=1451+35                  =5455+35                  =1585                  =18                  =122Hence, 1sinA1+cosA=122.

Page No 547:

Question 21:

If cot θ=158 then evaluate (1+sin θ) (1sin θ)(1+cos θ) (1cos θ)cot θ=158 then evaluate 1+sin θ 1sin θ1+cos θ 1cos θ.

Answer:

Given: cotθ=158Since, cotθ=BPP=8 and B=15Using Pythagoras theorem,P2+B2=H282+152=H2H2=64+225H