Rs Aggarwal 2021 2022 Solutions for Class 10 Maths Chapter 13 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among Class 10 students for Maths Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2021 2022 Book of Class 10 Maths Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2021 2022 Solutions. All Rs Aggarwal 2021 2022 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 617:

Answer:


sin4θ-cos4θ=sin2θ2-cos2θ2=sin2θ+cos2θsin2θ-cos2θ              a2-b2=a-ba+b
=1×1-cos2θ-cos2θ                           sin2θ+cos2θ=1=1-2cos2θ

Page No 617:

Answer:


We know
sin2θ+cos2θ=1
Squaring on both sides, we get
sin2θ+cos2θ2=1sin2θ2+cos2θ2+2sin2θcos2θ=1                 a+b2=a2+b2+2absin4θ+cos4θ=1-2sin2θcos2θ
sin4θ+cos4θ1-2sin2θcos2θ=1-2sin2θcos2θ1-2sin2θcos2θ=1

Page No 617:

Answer:

Ans

Page No 617:

Answer:


secA-cosAcotA+tanA=1cosA-cosAcosAsinA+sinAcosA=1-cos2AcosAcos2A+sin2AsinAcosA
=sin2AcosA×1sinAcosA               sin2θ+cos2θ=1=sinAcosA×1cosA=tanAsecA

Page No 617:

Answer:


1-sinθ1+sinθ=1-sinθ1+sinθ×1-sinθ1-sinθ=1-sinθ21-sin2θ                        a+ba-b=a2-b2
=1-sinθ2cos2θ               sin2θ+cos2θ=1=1-sinθcosθ2=1cosθ-sinθcosθ2=secθ-tanθ2

Page No 617:

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Page No 617:

Answer:


sec4θ-tan4θ=sec2θ2-tan2θ2=sec2θ-tan2θsec2θ+tan2θ            a2-b2=a-ba+b
=1×1+tan2θ+tan2θ                        1+tan2θ=sec2θ=1+2tan2θ

Page No 617:

Answer:

LHS=sinθ(cotθ+cosecθ)sinθ(cotθcosecθ)         =sinθ{(cotθcosecθ)(cotθ+cosecθ)(cotθ+cosecθ)(cotθcosecθ)}         =sinθ{2cosecθ(cot2θcosec2θ)}         =sinθ(2cosecθ1)    (cosec2θcot2θ=1)         =sinθ.2cosecθ         =sinθ×2×1sinθ         =2         =RHS

Page No 617:

Answer:


cosecA-sinAcosecA+sinA=1sinA-sinA1sinA+sinA=1-sin2AsinA1+sin2AsinA=1-sin2A1+sin2A
=1-sin2Acos2A1+sin2Acos2A           (Dividing numerator and denominator by cos2A)
=1cos2A-sin2Acos2A1cos2A+sin2Acos2A=sec2A-tan2Asec2A+tan2A                  secθ=1cosθ and tanθ=sinθcosθ

Page No 617:

Answer:


sin2θtanθ+cos2θcotθ+2sinθcosθ=sin3θcosθ+cos3θsinθ+2sinθcosθ=sin4θ+cos4θ+2sin2θcos2θsinθcosθ=sin2θ+cos2θ2sinθcosθ
=1sinθcosθ                         sin2θ+cos2θ=1=sin2θ+cos2θsinθcosθ=sin2θsinθcosθ+cos2θsinθcosθ
=sinθcosθ+cosθsinθ=tanθ+cotθ

Page No 617:

Answer:


tanθ+secθ-1tanθ+secθ+1=tanθ+secθ2-1                             a-ba+b=a2-b2=tan2θ+sec2θ+2tanθsecθ-1=2tan2θ+2tanθsecθ                         1+tan2θ=sec2θ   
=2tanθtanθ+secθ=2×sinθcosθ×sinθcosθ+1cosθ=2sinθ1+sinθcos2θ=2sinθ1+sinθ1-sin2θ                    sin2θ+cos2θ=1
=2sinθ1+sinθ1-sinθ1+sinθ=2sinθ1-sinθ

Page No 617:

Answer:


1+cotA+tanAsinA-cosA=sinA+cotAsinA+tanAsinA-cosA-cotAcosA-tanAcosA=sinA+cosAsinA×sinA+tanAsinA-cosA-cotAcosA-sinAcosA×cosA
=sinA+cosA+tanAsinA-cosA-cotAcosA-sinA=sinAtanA-cotAcosA

Page No 617:

Answer:

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Page No 617:

Answer:


sec2θ-sin2θ-2sin4θ2cos4θ-cos2θ=sec2θ-sin2θ1-2sin2θcos2θ2cos2θ-1=sec2θ-sin2θ1-2sin2θcos2θ21-sin2θ-1                cos2θ+sin2θ=1
=sec2θ-sin2θ1-2sin2θcos2θ1-2sin2θ=sec2θ-tan2θ    =1                                          1+tan2θ=sec2θ

Page No 617:

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Page No 617:

Answer:


tanA+sinAtanA-sinA=sinAcosA+sinAsinAcosA-sinA=sinA1cosA+1sinA1cosA-1=secA+1secA-1                               secA=1cosA

Page No 617:

Answer:


cotA-cosAcotA+cosA=cosAsinA-cosAcosAsinA+cosA=cosA1sinA-1cosA1sinA+1=cosecA-1cosecA+1                          cosecA=1sinA

Page No 617:

Answer:


1-sinθ1+sinθ=1-sinθ1+sinθ×1-sinθ1-sinθ=1-sinθ21+sinθ1-sinθ=1-sinθ21-sin2θ                   a+ba-b=a2-b2
=1-sinθ2cos2θ                sin2θ+cos2θ=1=1-sinθcosθ2=1cosθ-sinθcosθ2=secθ-tanθ2

Page No 617:

Answer:


1+cosθ1-cosθ=1+cosθ1-cosθ×1+cosθ1+cosθ=1+cosθ21-cosθ1+cosθ=1+cosθ21-cos2θ                   a+ba-b=a2-b2
=1+cosθ2sin2θ                sin2θ+cos2θ=1=1+cosθsinθ2=1sinθ+cosθsinθ2=cosecθ+cotθ2

Page No 617:

Answer:


secθ-tanθsecθ+tanθ=secθ-tanθsecθ+tanθ×secθ-tanθsecθ-tanθ=secθ-tanθ2secθ+tanθsecθ-tanθ=sec2θ+tan2θ-2secθtanθsec2θ-tan2θ            a-b2=a2+b2-2ab and a+ba-b=a2-b2
=1+tan2θ+tan2θ-2secθtanθ1             1+tan2θ=sec2θ=1+2tan2θ-2secθtanθ



Page No 618:

Answer:

ANS

Page No 618:

Answer:

LHS=cos3θ+sin3θcosθ+sinθ+cos3θsin3θcosθsinθ          =(cosθ+sinθ)(cos2θcosθsinθ+sin2θ)(cosθ+sinθ)+(cosθsinθ)(cos2θ+cosθsinθ+sin2θ)(cosθsinθ)          =(cos2θ+sin2θcosθsinθ)+(cos2θ+sin2θ+cosθsinθ)          =(1cosθsinθ)+(1+cosθsinθ)          =2          =RHS
Hence, LHS= RHS

Page No 618:

Answer:

LHS=1+cosθsin2θsinθ(1+cosθ)         =(1+cosθ)(1cos2θ)sinθ(1+cosθ)         =cosθ+cos2θsinθ(1+cosθ)         =cosθ(1+cosθ)sinθ(1+cosθ)        =cosθsinθ        =cotθ        =RHS
Hence, L.H.S. = R.H.S.

Page No 618:

Answer:

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Page No 618:

Answer:

ANS

Page No 618:

Answer:

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Page No 618:

Answer:

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Page No 618:

Answer:

LHS=cosθcosecθsinθsecθcosθ+sinθ         =cosθsinθsinθcosθcosθ+sinθ         =cos2θsin2θcosθsinθ(cosθ+sinθ)         =(cosθ+sinθ)(cosθsinθ)cosθsinθ(cosθ+sinθ)         =(cosθsinθ)cosθsinθ         =1sinθ1cosθ         =cosecθsecθ         =RHS         
Hence, LHS = RHS

Page No 618:

Answer:

LHS=(1+tanθ+cotθ)(sinθcosθ)         =sinθ+tanθsinθ+cotθsinθcosθtanθcosθcotθcosθ         =sinθ+tanθsinθ+cosθsinθ×sinθcosθsinθcosθ×cosθcotθcosθ         =sinθ+tanθsinθ+cosθcosθsinθcotθcosθ         =tanθsinθcotθcosθ         =sinθcosθ×1cosecθcosθsinθ×1secθ         =1cosecθ×1cosecθ×secθ1secθ×1secθ×cosecθ         =secθcosec2θcosecθsec2θ         =RHS

Hence, LHS = RHS

Page No 618:

Answer:

LHS=cot2θsecθ-11+sinθ+sec2θsinθ-11+secθ=cos2θsin2θ1cosθ-11+sinθ+1cos2θsinθ-11+1cosθ=cos2θsin2θ1-cosθcosθ1+sinθ+sinθ-1cos2θcosθ+1cosθ=cos2θ1-cosθsin2θ cosθ1+sinθ+sinθ-1cosθcosθ+1cos2θ=cosθ1-cosθ1-cos2θ1+sinθ+sinθ-1cosθcosθ+11-sin2θ=cosθ1-cosθ1-cosθ1+cosθ1+sinθ+-1-sinθcosθcosθ+11-sinθ1+sinθ=cosθ1+cosθ1+sinθ-cosθcosθ+11+sinθ=0=RHS

Page No 618:

Answer:

(i) cos2θ+cosθ=1LHS=cos2θ+cosθ         =1sin2θ+cosθ         =1(sin2θcosθ)        Since LHSRHS, this is not an identity.(ii) sin2θ+sinθ=1LHS=sin2θ+sinθ          =1cos2θ+sinθ          =1(cos2θsinθ)      Since LHS≠RHS, this is not an identity.(iii) tan2θ+sinθ=cos2θLHS=tan2θ+sinθ         =sin2θcos2θ+sinθ         =1cos2θcos2θ+sinθ         =sec2θ1+sinθSince LHSRHS, this is not an identity.



Page No 628:

Answer:

We have m2+n2=[acosθ+bsinθ2+asinθ-bcosθ2]                            = (a2cos2θ+b2sin2θ+2abcosθsinθ)+(a2sin2θ+b2cos2θ2absinθcosθ)                            = a2cos2θ+b2sin2θ+a2sin2θ+b2cos2θ                            =(a2cos2θ+a2sin2θ)+(b2cos2θ+b2sin2θ)                           =a2(cos2θ+sin2θ)+b2(cos2θ+sin2θ)                           =a2+b2           [sin2+cos2=1]Hence, m2+n2=a2+b2

Page No 628:

Answer:

We have x2y2=[asecθ+btanθ2-atanθ+bsecθ2]                            = (a2sec2θ+b2tan2θ+2absecθtanθ)(a2tan2θ+b2sec2θ+2abtanθsecθ)                            = a2sec2θ+b2tan2θa2tan2θb2sec2θ                            =(a2sec2θa2tan2θ)(b2sec2θb2tan2θ)                           =a2(sec2θtan2θ)b2(sec2θtan2θ)                           =a2b2            [sec2θtan2θ=1]Hence, x2y2=a2b2

Page No 628:

Answer:

We have (xasinθybcosθ)=1Squaring both side, we have:(xasinθybcosθ)2=(1)2(x2a2sin2θ+y2b2cos2θ2xa×ybsinθcosθ)=1           ...(i)Again, (xacosθ+ybsinθ)=1Squaring both side, we get:(xacosθ+ybsinθ)2=(1)2(x2a2cos2θ+y2b2sin2θ+2xa×ybsinθcosθ)=1         ...(ii)Now, adding (i) and (ii), we get:(x2a2sin2θ+y2b2cos2θ2xa×ybsinθcosθ)+(x2a2cos2θ+y2b2sin2θ+2xa×ybsinθcosθ)=2x2a2sin2θ+y2b2cos2θ+x2a2cos2θ+y2b2sin2θ=2(x2a2sin2θ+x2a2cos2θ)+(y2b2cos2θ+y2b2sin2θ)=2x2a2(sin2θ+cos2θ)+y2b2(cos2θ+sin2θ)=2x2a2+y2b2=2           [sin2θ+cos2θ=1]∴ x2a2+y2b2=2

Page No 628:

Answer:

We have x=acos3θ          =>xa=cos3θ      ...(i)Again, y=bsin3θ          =>yb=sin3θ       ...(ii)Now, LHS=(xa)23+(yb)23        =(cos3θ)23+(sin3θ)23         [From (i) and (ii)]         =cos2θ+sin2θ        =1 Hence, LHS= RHS

Page No 628:

Answer:


x=bsec3θsec3θ=xbsecθ=xb13              .....1
Also,
y=atan3θtan3θ=yatanθ=ya13              .....2
We know
sec2θ-tan2θ=1xb132-ya132=1         Using 1 and 2xb23-ya23=1

Page No 628:

Answer:

We have (tanθ+sinθ)=m and (tanθsinθ)=nNow, LHS=(m2n2)2         =[(tanθ+sinθ)2(tanθsinθ)2]2        =[(tan2θ+sin2θ+2tanθsinθ)(tan2θ+sin2θ2tanθsinθ)]2        =[(tan2θ+sin2θ+2tanθsinθtan2θsin2θ+2tanθsinθ)]2        =(4tanθsinθ)2        =16tan2θsin2θ        =16sin2θcos2θsin2θ        =16(1cos2θ)sin2θcos2θ        =16[tan2θ(1cos2θ)]        =16(tan2θtan2θcos2θ)        =16(tan2θsin2θcos2θ×cos2θ)        =16(tan2θsin2θ)        =16(tanθ+sinθ)(tanθsinθ)        =16mn                                         [(tanθ+sinθ)(tanθsinθ)=mn](m2n2)(m2n2)2=16mn

Page No 628:

Answer:

We have (cotθ+tanθ)=m and (secθcosθ)=nNow, m2n=[(cotθ+tanθ)2(secθcosθ)]=1tanθ+tanθ21cosθ-cosθ=(1+tan2θ)2tan2θ×(1cos2θ)cosθ=sec4θtan2θ×sin2θcosθ=sec4θsin2θcos2θ×sin2θcosθ=cos2θ×sec4θcosθ=cosθsec4θ=1secθ×sec4θ=sec3θ(m2n)23=(sec3θ)23=sec2θ              

Again, mn2=[(cotθ+tanθ)(secθcosθ)2]=[(1tanθ+tanθ).(1cosθcosθ)2]=(1+tan2θ)tanθ×(1cos2θ)2cos2θ=sec2θtanθ×sin4θcos2θ=sec2θsinθcosθ×sin4θcos2θ=sec2θ×sin3θcosθ=1cos2θ×sec3θcosθ=tan3θ(mn2)23=(tan3θ)23=tan2θNow, (m2n)23(mn2)23=sec2θtan2θ=1=RHSHence proved.

Page No 628:

Answer:

We have (cosecθ-sinθ)=a3=> a3=(1sinθsinθ)=> a3=(1sin2θ)sinθ=cos2θsinθa=cos23θsin13θAgain, (secθcosθ)=b3=>b3=(1cosθcosθ)=(1cos2θ)cosθ=sin2θcosθ b=sin23θcos13θNow, LHS=a2b2(a2+b2)  =a4b2+a2b4=a3(ab2)+(a2b)b3=cos2θsinθ×cos23θsin13θ×sin43θcos23θ+cos43θsin23θ×sin23θcos13θ×sin2θcosθ =cos2θsinθ×sinθ+cosθ×sin2θcosθ=cos2θ+sin2θ=1= RHSHence proved.



Page No 629:

Answer:


Given:

x = secA + sinA       .....(1)

y = secA – sinA       .....(2)

Adding (1) and (2), we get

x+y=secA+sinA+secA-sinA2secA=x+ysecA=x+y21secA=2x+y
cosA=2x+y            .....3

Subtracting (2) from (1), we get

x-y=secA+sinA-secA+sinA2sinA=x-ysinA=x-y2             .....4

We know

cos2A+sin2A=12x+y2+x-y22=1           Using 3 and 4

Page No 629:

Answer:

LHS=mn+nm=mn+nm=m+nmn=cosθ-sinθ+cosθ+sinθcosθ-sinθcosθ+sinθ
=2cosθcos2θ-sin2θ=2cosθcosθcos2θ-sin2θcosθ=2cos2θcos2θ-sin2θcos2θ=21-tan2θ=RHS

Page No 629:

Answer:


cosθ-sinθ=2sinθ
Squaring on both sides, we get
cosθ-sinθ2=2sinθ2cos2θ+sin2θ-2sinθcosθ=2sin2θcos2θ-sin2θ=2sinθcosθcosθ-sinθcosθ+sinθ=2sinθcosθ                  a2-b2=a-ba+b
2sinθcosθ+sinθ=2sinθcosθ              cosθ-sinθ=2sinθcosθ+sinθ=2sinθcosθ2sinθcosθ+sinθ=2cosθ

Page No 629:

Answer:


secθ-tanθ=2tanθ
Squaring on both sides, we get
secθ-tanθ2=2tanθ2sec2θ+tan2θ-2secθtanθ=2tan2θsec2θ-tan2θ=2secθtanθsecθ-tanθsecθ+tanθ=2secθtanθ          a2-b2=a-ba+b
2tanθsecθ+tanθ=2secθtanθ            secθ-tanθ=2tanθsecθ+tanθ=2secθtanθ2tanθsecθ+tanθ=2secθ

Page No 629:

Answer:


Given: secθ+tanθ=p            .....1
We know
sec2θ-tan2θ=1secθ-tanθsecθ+tanθ=1secθ-tanθp=1                            From 1secθ-tanθ=1p               .....2
Adding (1) and (2), we get
secθ+tanθ+secθ-tanθ=p+1p2secθ=p2+1psecθ=p2+12p1secθ=2pp2+1
cosθ=2pp2+1            .....3
Subtracting (2) from (1), we get
secθ+tanθ-secθ+tanθ=p-1p2tanθ=p2-1ptanθ=p2-12p             .....4
Now,
sinθ=tanθ×cosθsinθ=p2-12p×2pp2+1           Using 3 and 4sinθ=p2-1p2+1

Page No 629:

Answer:


Given: cosecA+cotA=m         .....1
We know
cosec2A-cot2A=1cosecA-cotAcosecA+cotA=1             a2-b2=a-ba+bcosecA-cotAm=1                                      From 1cosecA-cotA=1m         .....2
Adding (1) and (2), we get
cosecA+cotA+cosecA-cotA=m+1m2cosecA=m2+1mcosecA=m2+12m1cosecA=2mm2+1
sinA=2mm2+1          .....3
Subtracting (2) from (1), we get
cosecA+cotA-cosecA+cotA=m-1m2cotA=m2-1mcotA=m2-12m          .....4
Now,
cosA=sinA×cotAcosA=2mm2+1×m2-12m        From 3 and 4cosA=m2-1m2+1

Page No 629:

Answer:


Given: secA-tanA=x        .....1
We know
sec2A-tan2A=1secA+tanAsecA-tanA=1            a2-b2=a-ba+bsecA+tanAx=1                                  From 1secA+tanA=1x               .....2
Adding (1) and (2), we get
secA-tanA+secA+tanA=x+1x2secA=x2+1xsecA=x2+12x            .....3
Subtracting (1) from (2), we get
secA+tanA-secA+tanA=1x-x2tanA=1-x2xtanA=1-x22x            .....4
Dividing (3) by (4), we get
1+x22x1-x22x=secAtanA1+x21-x2=1cosAsinAcosA1+x21-x2=1sinA1+x21-x2=cosecA



Page No 630:

Answer:

(b) 1
  sinA+sin2A=1=>sinA=1sin2A =>sinA=cos2A   (1sin2A)=>sin2A=cos4A   (Squaring both sides)=>1cos2A=cos4A=> cos4A+cos2A=1

Page No 630:

Answer:


1+tan2A1+cot2A=sec2Acosec2A=sin2Acos2A                 secθ=1cosθ and cosecθ=1sinθ=tan2A

Hence, the correct answer is option (c).



Page No 631:

Answer:

(b) cos A

(secA+tanA)(1sinA)          =(1cosA+sinAcosA)(1sinA)         =(1+sinAcosA)(1sinA)         =(1sin2AcosA)         =(cos2AcosA)         =cosA

Page No 631:

Answer:


1+tanθ+secθ1+cotθ-cosecθ=1+cotθ-cosecθ+tanθ+tanθcotθ-tanθcosecθ+secθ+secθcotθ-secθcosecθ=1+cosθsinθ-1sinθ+sinθcosθ+tanθ×1tanθ-sinθcosθ×1sinθ+1cosθ+1cosθ×cosθsinθ-1cosθsinθ
=1+cosθsinθ-1sinθ+sinθcosθ+1-1cosθ+1cosθ+1sinθ-1cosθsinθ=2+cosθsinθ+sinθcosθ-1cosθsinθ=2+cos2θ+sin2θcosθsinθ-1cosθsinθ
=2+1cosθsinθ-1cosθsinθ            cos2θ+sin2θ=1=2

Hence, the correct answer is option (d).

Page No 631:

Answer:


sinθ-cosθ=0sinθ=cosθsinθcosθ=1tanθ=tan45°
θ=45°sin4θ+cos4θ=sin445°+cos445°=124+124=14+14
=24=12

Hence, the correct answer is option (b).

Page No 631:

Answer:


cos9α=sinαcos9α=cos90°-α9α=90°-α10α=90°
5α=90°2=45°tan5α=tan45°=1

Thus, the value of tan5α is 1.

Hence, the correct answer is option (c).

Page No 631:

Answer:


tan2θ-sec2θcot2θ-cosec2θ=-1-1                   1+tan2θ=sec2θ and 1+cot2θ=cosec2θ=1

Hence, the correct answer is option (d).

Page No 631:

Answer:

(b) (sec A − tan A)

1sinA1+sinA=(1sinA)(1+sinA)×(1sinA)(1sinA)          [Multiplying the denominator and numerator by (1sinA)]=(1sinA)1sin2A=(1+sinA)cos2A=(1sinA)cosA=1cosAsinAcosA=secAtanA

Page No 631:

Answer:

1+cosA1-cosA=1+cosA1-cosA×1+cosA1+cosA=1+cosA1-cos2A=1+cosAsin2A=1+cosAsinA=1sinA+cosAsinA=cosecA+cotA.
Hence, the correct answer is option B.

Page No 631:

Answer:

(b) 1-cosθ1+cosθ

(cosecθcotθ)2=(1sinθcosθsinθ)2=(1cosθsinθ)2=(1cosθ)2sin2θ=(1cosθ)2(1cos2θ)=(1cosθ)2(1+cosθ)(1cosθ)=(1cosθ)(1+cosθ)

Page No 631:

Answer:

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)2 = 52
⇒ tan2 θ + cot2 θ + 2 tan θ cot θ = 25
⇒ tan2 θ + cot2 θ + 2 = 25       [∵ tan θ = 1cot θ]
⇒ tan2 θ + cot2 θ = 25 − 2 = 23

Page No 631:

Answer:

(c) b+ab-a
Given: tanθ=ab Now,(cosθ+sinθ)(cosθsinθ)=(1+tanθ)(1tanθ)     [Dividing the numerator and denominator by cosθ]=(1+ab)(1ab)=(b+ab)(bab)=(b+a)(ba)

Page No 631:

Answer:

(d) 34
=cosec2θ - sec2θcosec2θ + sec2θ = sin2θ1sin2θ-1cos2θsin2θ1sin2θ +1cos2θ      [Multiplying  the numerator and denominator by sin2θ]= 1 - tan2θ1 + tan2θ
= 1-171 +17 = 68 = 34



Page No 632:

Answer:


tanθ=8151tanθ=158cotθ=158          .....1
Now,
cosec2θ=1+cot2θcosec2θ=1+1582             Using 1cosec2θ=1+22564=28964cosecθ=28964=178

Hence, the correct answer is option (a).

Page No 632:

Answer:


5tanθ=35×sinθcosθ=35sinθ=3cosθ      .....1
5sinθ-cosθ5sinθ+cosθ=3cosθ-cosθ3cosθ+cosθ              Using 1=2cosθ4cosθ=12

Hence, the correct answer is option (c).

Page No 632:

Answer:


In ∆ABC,

∠A + ∠B + ∠C = 180º                (Angle sum property of triangle)

⇒ ∠A + ∠B + 90º = 180º            (∠C = 90º)

⇒ ∠A + ∠B = 180º − 90º = 90º

∴ cos(A + B) = cos90º = 0

Thus, the value of cos(A + B) is 0.

Hence, the correct answer is option (a).

Page No 632:

Answer:

(a) 1
 cosA+cos2A=1=>cosA=1cos2A=>cosA=sin2A   (1cos2A=sin2)=>cos2A=sin4A   (Squaring both sides)=>1sin2A=sin4A=> sin4A+sin2A=1

Page No 632:

Answer:

(d) 9

We have 5sinθ + 3cosθ5sinθ - 3cosθ.

Dividing the numerator and denominator of the given expression by sin θ, we get:
 1sinθ5sinθ + 3cosθ1sinθ5sinθ - 3cosθ

= 5 + 3cot θ5 - 3cot θ

= 5 + 45 - 4 = 9              [∵ 3 cot θ = 4]

Page No 632:

Answer:

(a) 12
Given: 2x = sec A and 2x = tan A
Also, we can deduce that x = sec A2 and 1x = tan A2.
So, substituting the values of x and 1x in the given expression, we get:
2x2-1x2 = 2sec A22 - tan A22
= 2sec2A4 - tan2A4
= 24sec2A -tan2A
= 12                  [By using the identity: sec2θ- tan2θ = 1]

Page No 632:

Answer:

(c) 13
Given: 3x = cosec θ and 3x = cot θ
Also, we can deduce that x = cosec θ3 and 1x = cot θ3.
So, substituting the values of x and 1x in the given expression, we get:
3x2-1x2 = 3cosec θ32 - cot θ32
= 3cosec2θ9 - cot2θ9
= 39cosec2θ -cot2θ
= 13       [By using the identity: cosec2θ- cot2θ = 1]

Page No 632:

Answer:

(d) 2

Let us first draw a right ABC right angled at B and A=θ.
Given: tan θ = 3
But tan θ = BCAB
So, BCAB = 31
Thus, BC = 3k and AB = k


Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
⇒ AC2 = (3 k)2 + (k)2
⇒ AC2= 4k2
⇒ AC = 2k
∴ sec θ = ACAB = 2kk = 21



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