Rs Aggarwal 2020 2021 Solutions for Class 10 Maths Chapter 10 Trignometric Ratios are provided here with simple step-by-step explanations. These solutions for Trignometric Ratios are extremely popular among Class 10 students for Maths Trignometric Ratios Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2020 2021 Book of Class 10 Maths Chapter 10 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 546:

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:

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:

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:

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:

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:

Answer:

We have sinθ=a2-b2a2+b2,

As,

cos2θ=1-sin2θ=1-a2-b2a2+b22=11-a2-b22a2+b22=a2+b22-a2-b22a2+b22=a2+b2-a2-b2a2+b2+a2-b2a2+b22
=a2+b2-a2+b2a2+b2+a2-b2a2+b22=2b22a2a2+b22cos2θ=4a2b2a2+b22cosθ=4a2b2a2+b22cosθ=2aba2+b2

Also,

tanθ=sinθcosθ=a2-b2a2+b22aba2+b2=a2-b22ab

Now,

cosecθ=1sinθ=1a2-b2a2+b2=a2+b2a2-b2

Also,

secθ=1cosθ=12aba2+b2=a2+b22ab

And,

cotθ=1tanθ=1a2-b22ab=2aba2-b2

Page No 546:

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:

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=32cos2θ-sin2θ=322-122                   =34-14=24                   =12Hence, cos2θ-sin2θ=12.

Page No 546:

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:

Answer:

LHS=secθ+tanθ=1cosθ+sinθcosθ=1+sinθcosθ=1+sinθ1-sin2θ=1+ab1-ab2
=11+ab11-a2b2=b+abb2-a2b2=b+abb2-a2b=b+ab+ab-a
=b+ab+ab-a=b+ab-a=b+ab-a=RHS



Page No 547:

Answer:

It is given that tan θ = ab.

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

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

 Hence proved.

Page No 547:

Answer:

Given: sinθ=1213Since, sinθ=PHP=12 and H=13Using Pythagoras theorem,P2+B2=H2122+B2=132B2=169-144B2=25B=5Therefore,cosθ=BH=513Now,2sinθ-3cosθ4sinθ-9cosθ=21213-351341213-9513                        =2413-15134813-4513                        =24-1548-45                        =93                        =3Hence, 2sinθ-3cosθ4sinθ-9cosθ=3.

Page No 547:

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+2×5-25-2                             =2+5-25-4                             =2+5-2                             =5Hence, cosθsinθ+sinθ1+cosθ=5.

Page No 547:

Answer:

LHS=3cosα-4cos3α=cosα3-4cos2α=1-sin2α3-41-sin2α=1-1223-41-122=11-143-411-14=343-434=343-3=340=0=RHS

Page No 547:

Answer:

It is given that cot θ = 23.

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

Page No 547:

Answer:

It is given that sec θ = 178.

Let us consider a right ABC right angled at B and C=θ.
We know that cos θ = 1sec 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 θ  = ABBC = 158 and sin θ = ABAC = 15k17k= 1517

The given expression is 3 - 4sin2θ4cos2θ- 3 = 3 - tan2θ1 - 3tan2θ.
 
 Substituting the values in the above expression, we get:
 LHS= 3 - 41517248172 - 3 = 3 - 900289256289- 3 = 867-900256-867= -33-611=33611

RHS = 3-15821-31582=3-225641-67564=192-22564-675=-33-611=33611

∴ LHS = RHS
Hence proved.

Page No 547:

Answer:

Let us consider a right ABC right angled at B and C=θ.
Now, we know that tan θABBC = 2021

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 θ = ABAC = 2029 and cos θ = BCAC = 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 = RHS

Hence proved.

Page No 547:

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:

Answer:

LHS=cosec2θ-cot2θsec2θ-1=1tan2θ=cot2θ=cotθ=cosec2θ-1=1sinθ2-1=1342-1=432-1=169-1=16-99=79=73=RHS

Page No 547:

Answer:

(i)
 ​LHS=secθ-cosecθsecθ+cosecθ=1cosθ-1sinθ1cosθ+1sinθ=sinθ-cosθsinθ cosθsinθ+cosθsinθ cosθ=sinθ-cosθsinθsinθ+cosθsinθ=sinθsinθ-cosθsinθsinθsinθ+cosθsinθ=1-cotθ1+cotθ=1-341+34
=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,1-sinA1+cosA=1-451+35                  =5-455+35                  =1585                  =18                  =122Hence, 1-sinA1+cosA=122.

Page No 547:

Answer:

Given: cotθ=158Since, cotθ=BPP=8 and B=15Using Pythagoras theorem,P2+B2=H282+152=H2H2=64+225H2=289H=17Therefore,sinθ=PH=817cosθ=BH=1517Now,1+sinθ1-sinθ1+cosθ1-cosθ=1-sin2θ1-cos2θ                               =cos2θsin2θ             sin2θ+cos2θ=1                               =cot2θ                               =1582                               =22564Hence, 1+sinθ1-sinθ1+cosθ1-cosθ=22564.

Page No 547:

Answer:

We have,tanA=1sinAcosA=1sinA=cosAsinA-cosA=0Squaring both sides, we getsinA-cosA2=0sin2A+cos2A-2sinA·cosA=01-2sinA·cosA=02sinA·cosA=1

Page No 547:

Answer:

Given: In ABC,AC=17 cmsinθ=817Since, sinθ=PHP=8 and H=17Using Pythagoras theorem,P2+B2=H282+B2=172B2=289-64B2=225B=15Therefore,AB= 8 cm and BC=15 cmTherefore,i Area of rectangle ABCD=AB×BC                                       =8×15                                       =120 cm2ii Perimeter of rectangle ABCD=2AB+BC                                               =28+15                                               =223                                               =46 cm

Page No 547:

Answer:

LHS=2x+y2+x-y22-1=2cosecA+cosA+cosecA-cosA2+cosecA+cosA-cosecA-cosA22-1=2cosecA+cosA+cosecA-cosA2+cosecA+cosA-cosecA+cosA22-1=22cosecA2+2cosA22-1
=1cosecA2+cosA2-1=sinA2+cosA2-1=sin2A+cos2A-1=1-1=0=RHS



Page No 548:

Answer:

LHS=x-yx+y2+x-y22=cotA+cosA-cotA-cosAcotA+cosA+cotA-cosA2+cotA+cosA-cotA-cosA22=cotA+cosA-cotA+cosAcotA+cosA+cotA-cosA2+cotA+cosA-cotA+cosA22=2cosA2cotA2+2cosA22=cosAcosAsinA2+cosA2=sinA cosAcosA2+cosA2=sinA2+cosA2=sin2A+cos2A=1=RHS

Page No 548:

Answer:


In PQR, Q=90°,

Using Pythagoras theorem, we get

PQ=PR2-QR2=x+22-x2=x2+4x+4-x2=4x+1=2x+1

Now,

i x+1cotϕ=x+1×QRPQ=x+1×x2x+1=x2

ii x3+x2tanθ=x2x+1×QRPQ=xx+1×x2x+1=x22

iii cosθ=PQPR=2x+1x+2

Page No 548:

Answer:

Given:cotA+1cotA=2cotA+1cotA=2Squaring both sides, we getcotA+1cotA2=22cot2A+1cotA2+2cotA1cotA=4cot2A+1cot2A+2=4cot2A+1cot2A=4-2cot2A+1cot2A=2Hence, the value of cot2A+1cot2A is 2.

Page No 548:

Answer:

Given:3tanθ=3sinθ3tanθ=3sinθ3sinθcosθ=3sinθ3sinθcosθ-3sinθ=03sinθ-3sinθcosθcosθ=03sinθ-3sinθcosθ=03sinθ1-3cosθ=0sinθ=0 or 1-3cosθ=0sinθ=0 or cosθ=13sinθ=0 or cos2θ=13sinθ=0 or 1-cos2θ=1-13sinθ=0 or sin2θ=23sinθ=0 or sinθ=23Hence, the value of sinθ is 0 or 23.

Page No 548:

Answer:


In ABC, C = 90°
sinA = BCAB and
sinB = ACAB

As, sinA = sinB
BCAB = ACAB
BC = AC
So, A = B             (Angles opposite to equal sides are equal)

Page No 548:

Answer:



In ABC, C=90°,

tanA=BCAC andtanB=ACBC

As, tanA=tanB
BCAC=ACBCBC2=AC2BC=ACSo, A=B             Angles opposite to equal sides are equal



Page No 555:

Answer:

Given: tanθ=815Since, tanθ=PBP=8 and B=15Using Pythagoras theorem,P2+B2=H282+152=H2H2=64+225H2=289H=17Therefore,cosecθ=HP=178

Hence, the correct option is (c).

Page No 555:

Answer:

Given: tanθ=31Since, tanθ=PBP=3 and B=1Using Pythagoras theorem,P2+B2=H232+12=H2H2=3+1H2=4H=2Therefore,secθ=HB=21=2

Hence, the correct option is (a).

Page No 555:

Answer:

Given: cosecθ=101Since, cosecθ=HPP=1 and H=10Using Pythagoras theorem,P2+B2=H212+B2=102B2=10-1B2=9B=3Therefore,secθ=HB=103

Hence, the correct option is (d).

Page No 555:

Answer:

Given: secθ=257Since, secθ=HBB=7 and H=25Using Pythagoras theorem,P2+B2=H2P2+72=252P2=625-49P2=576P=24Therefore,sinθ=PH=2425

Hence, the correct option is (b).

Page No 555:

Answer:

Given: sinθ=12Since, sinθ=PHP=1 and H=2Using Pythagoras theorem,P2+B2=H212+B2=22B2=4-1B2=3B=3Therefore,cotθ=BP=31

Hence, the correct option is (c).

Page No 555:

Answer:

Given: cosθ=45Since, cosθ=BHB=4 and H=5Using Pythagoras theorem,P2+B2=H2P2+42=52P2=25-16P2=9P=3Therefore,tanθ=PB=34

Hence, the correct option is (a).



Page No 556:

Answer:

Given: tanθ=43Since, tanθ=PBP=4 and B=3Using Pythagoras theorem,P2+B2=H242+32=H2H2=16+9H2=25H=5Therefore,sinθ=PH=45cosθ=BH=35Now,sinθ+cosθ=45+35                 =75

Hence, the correct option is (c).

Page No 556:

Answer:

Given:tanθ+cotθ=5tanθ+cotθ=5Squaring both sides, we gettanθ+cotθ2=52tan2θ+cot2θ+2cotθtanθ=25tan2θ+cot2θ+21tanθtanθ=25           cotθ=1tanθtan2θ+cot2θ+2=25tan2θ+cot2θ=23Hence, the correct option is d.

Page No 556:

Answer:

Given:cosθ+secθ=52cosθ+secθ=52Squaring both sides, we getcosθ+secθ2=522cos2θ+sec2θ+2cosθsecθ=254cos2θ+sec2θ+2cosθ1cosθ=254            secθ=1cosθcos2θ+sec2θ+2=254cos2θ+sec2θ=254-2cos2θ+sec2θ=25-84cos2θ+sec2θ=174Hence, the correct option is a.

Page No 556:

Answer:

Given: 4tanθ=3tanθ=34Since, tanθ=PBP=3 and B=4Using Pythagoras theorem,P2+B2=H232+42=H2H2=9+16H2=25H=5Therefore,sinθ=PH=35cosθ=BH=45cos2θ-sin2θ=452-352                   =1625-925                   =16-925                   =725Hence, the correct option is b.

Page No 556:

Answer:

Given: 4cotθ=3cotθ=34Since, cotθ=BPP=4 and B=3Using Pythagoras theorem,P2+B2=H242+32=H2H2=16+9H2=25H=5Therefore,sinθ=PH=45cosθ=BH=35sinθ-cosθsinθ+cosθ=45-3545+35                     =4-354+35                     =17Hence, the correct option is c.

Page No 556:

Answer:

Given: 3cosθ=2cosθ=23Since, cosθ=BHB=2 and H=3Using Pythagoras theorem,P2+B2=H2P2+22=32P2=9-4P2=5P=5Therefore,secθ=HB=32tanθ=PB=52Now,2sec2θ+2tan2θ-7=2322+2522-7                               =294+254-7                               =92+52-7                               =9+52-7                               =142-7                               =7-7                               =0Hence, the correct option is a.

Page No 556:

Answer:

Given: secθ+tanθ+1=0secθ+tanθ+1=0secθ+tanθ=-1Multiplying and dividing LHS by secθ-tanθ, we getsecθ+tanθ×secθ-tanθsecθ-tanθ=-1sec2θ-tan2θsecθ-tanθ=-11+tan2θ-tan2θsecθ-tanθ=-1           sec2θ=1+tan2θ1secθ-tanθ=-1secθ-tanθ=-1Hence, the correct option is b.

Page No 556:

Answer:

Given: cosA+cos2A=1cosA+cos2A=1cosA=1-cos2AcosA=sin2A          sin2A+cos2A=1Now,sin2A+sin4A=sin2A+sin2A2                   =cosA+cosA2                   =cosA+cos2A                   =1Hence, the correct option is c.

Page No 556:

Answer:

Given: sinθ=32Since, sinθ=PHP=3 and H=2Using Pythagoras theorem,P2+B2=H232+B2=22B2=4-3B2=1B=1Therefore,cosecθ=HP=23cotθ=BP=13Now,cosecθ+cotθ=23+13                      =2+13                      =33                      =3Hence, the correct option is d.

Page No 556:

Answer:

Given:3tanθ=3sinθ3tanθ=3sinθ3sinθcosθ=3sinθ3sinθcosθ-3sinθ=03sinθ-3sinθcosθcosθ=03sinθ-3sinθcosθ=03sinθ1-3cosθ=0sinθ=0 or 1-3cosθ=0sinθ=0 or cosθ=13sinθ=0 or cos2θ=13sinθ=0 or 1-cos2θ=1-13sinθ=0 or sin2θ=23sinθ=0 or sinθ=23For sinθ=0,sin2θ=01-sin2θ=1-0cos2θ=1Thus, sin2θ-cos2θ=-1For sinθ=23,sin2θ=231-sin2θ=1-23cos2θ=13Thus, sin2θ-cos2θ=23-13=13 

Hence, the correct option is (a).
 



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