Rs Aggarwal 2020 2021 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 2020 2021 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 2020 2021 Solutions. All Rs Aggarwal 2020 2021 Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

Page No 617:

Question 1:

sin4θ – cos4θ = 1 – 2cos2θ

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:

Question 2:

sin4θ+cos4θ1-2sin2θ cos2θ=1

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:

Question 3:

1+cot2θ1+cosecθ=cosecθ

Answer:

Ans

Page No 617:

Question 4:

(sec A – cos A) (cot A + tan A) = sec A tan A

Answer:


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

Page No 617:

Question 5:

1-sin θ1+sin θ=sec θ-tan θ2

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:

Question 6:

sin A(1 + tan A) + cos A(1 + cot A) = sec A + cosec A

Answer:

Ans

Page No 617:

Question 7:

sec4θ – tan4θ = 1 + 2 tan2θ

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:

Question 8:

sinθ(cotθ+cosecθ)-sinθ(cotθ-cosecθ)=2

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:

Question 9:

cosec A-sin Acosec A+sin A=sec2A-tan2Asec2A+tan2A

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:

Question 10:

sin2θ tanθ + cos2θ cot θ + 2sin θ cos θ = tan θ + cot θ

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:

Question 11:

tan θ+sec θ-1tan θ+sec θ+1=2sin θ1-sin θ

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:

Question 12:

(1 + cot A + tan A) (sin A – cos A) = sin A tan A – cot A cos A

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:

Question 13:

secθ+tanθsecθ-tanθ=1+sinθcosθ2

Answer:

Ans

Page No 617:

Question 14:

sec2θ-sin2θ-2sin4θ2cos4θ-cos2θ=1

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:

Question 15:

sinA(secA+tanA-1)+cosA(cosecA+cotA-1)=1

Answer:

Ans

Page No 617:

Question 16:

tan A+sin Atan A-sin A=sec A+1sec A1

Answer:


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

Page No 617:

Question 17:

cot A-cos Acot A+cos A=cosec A-1cosec A+1

Answer:


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

Page No 617:

Question 18:

1-sin θ1+sin θ=sec θ-tan θ2

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:

Question 19:

1+cos θ1-cos θ=cosec θ+cot θ2

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:

Question 20:

sec θ-tan θsec θ+tan θ=1+2tan2θ-2sec θ tan θ

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:

Question 21:


1+tan2θ1+secθ=secθ

Answer:

ANS

Page No 618:

Question 22:

cos3θ+sin3θcosθ+sinθ+cos3θ-sin3θcosθ-sinθ=2

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:

Question 23:

1+cosθ-sin2θsinθ(1+cosθ)=cotθ

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:

Question 24:

cosecθ+cotθcosecθ-cotθ=(cosecθ+cotθ)2=1+2cot2θ+2cosecθ cotθ

Answer:

ANS

Page No 618:

Question 25:

secθ+tanθsecθ-tanθ=(secθ+tanθ)2=1+2tan2θ+2secθ tanθ

Answer:

ANS

Page No 618:

Question 26:

1+cosθ+sinθ1+cosθ-sinθ=1+sinθcosθ

Answer:

Ans

Page No 618:

Question 27:

sinθ+1-cosθcosθ-1+sinθ=1+sinθcosθ

Answer:

Ans

Page No 618:

Question 28:

cosθ cosecθ-sinθ secθcosθ+sinθ=cosecθ-secθ

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:

Question 29:

(1+tanθ+cotθ)(sinθ-cosθ)=secθcosec2θ-cosecθsec2θ

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:

Question 30:

Prove that cot2θsecθ-11+sinθ+sec2θsinθ-11+secθ=0

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:

Question 31:

Show that none of the following is an identity:
(i) cos2θ + cos θ = 1
(ii) sin2θ + sin θ = 2
(iii) tan2θ + sin θ = cos2θ

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:

Question 1:

If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that (m2 + n2) = (a2 + b2).

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:

Question 2:

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x2y2) = (a2b2).

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:

Question 3:

If xasinθ-ybcosθ=1 and xacosθ+ybsinθ=1, prove that x2a2+y2b2=2.

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:

Question 4:

If x = a cos3θ and y = b sin3θ, prove that xa2/3+yb2/3=1.

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:

Question 5:

If x = b sec3θ and y = a tan3θ, prove that xb2/3-ya2/3=1.

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:

Question 6:

If (tan θ + sin θ) = m and (tan θ − sin θ) = n, prove that (m2 n2)2 = 16mn.

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:

Question 7:

If (cot θ + tan θ) = m and (sec θ − cos θ) = n, prove that (m2n)2/3 − (mn2)2/3 = 1.

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:

Question 8:

If (cosec θ − sin θ) = a3 and (sec θ − cos θ) = b3, prove that a2b2(a2+b2)=1.

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:

Question 9:

If x = (sec A + sin A) and y = (sec A – sin A), prove that 2x+y2+x-y22=1.

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:

Question 10:

If m=cosθ-sinθ and n=cosθ+sinθ, then show that mn+nm=21-tan2θ.

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:

Question 11:

If (cos θ – sin θ) = 2 sin θ then prove that (cos θ + sin θ) = 2 cos θ.

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:

Question 12:

If (sec θ – tan θ) = 2 tan θ then prove that (sec θ + tan θ) = 2 sec θ.

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:

Question 13:

If (sec θ + tan θ ) = p then show that (sec θ – tan θ) = 1p.
Hence, show that cos θ = 2pp2+1and sin θ=p2-1p2+1.

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:

Question 14:

If (cosec A + cot A) = m then prove that m2-1m2+1=cos θ.

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:

Question 15:

If (sec A – tan A) = x then prove that 1+x21-x2= cosec A.

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:

Question 1:

If sin A + sin2 A = 1, then (cos2 A + cos4 A) = ?
(a) 12
(b) 1
(c) 2
(d) 3

Answer:

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

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Question 2:

1+tan2A1+cot2A=?
(a) –1
(b) sec2A
(c) tan2A
(d) cot2A

Answer:


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

Hence, the correct answer is option (c).



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Question 3:

(sec A + tan A) (1 − sin A) = ?
(a) sin A
(b) cos A
(c) sec A
(d) cosec A

Answer:

(b) cos A

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

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Question 4:

(1 + tan θ + sec θ) (1 + cot θ – cosec θ) = ?
(a) –1
(b) 0
(c) 1
(d) 2

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).

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Question 5:

If sin θ – cos θ = 0 then the value of (sin4θ + cos4θ) is 
(a) 14

(b) 12

(c) 34

(d) 1

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).

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Question 6:

If cos 9α = sin α and 9α < 90° then the value of tan 5α is

(a) 13

(b) 3

(c) 1

(d) 0

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).

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Question 7:

The value of tan2θ-sec2θcot2θ-cosec2θ is
(a) –1

(b) 12

(c) -12

(d) 1

Answer:


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

Hence, the correct answer is option (d).

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Question 8:

1-sin A1+sin A=?
(a) (sec A + tan A)
(b) (sec A − tan A)
(c) sec A tan A
(d) None to these

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

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Question 9:

1+cos A1-cos A=?
(a) cosec A – cot A
(b) cosec A + cot A
(c) cosec A cot A
(d) none of these

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.

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Question 10:

(cosec θ − cot θ)2 = ?
(a) 1+cosθ1-cosθ
(b) 1-cosθ1+cosθ
(c)  1+sinθ1-sinθ
(d) 1-sinθ1+sinθ

Answer:

(b) 1-cosθ1+cosθ

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

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Question 11:

If (tan θ + cot θ) = 5, then (tan2 θ + cot2 θ) = ?

(a) 23
(b) 24
(c) 25
(d) 27

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

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Question 12:

If tan θ = ab, then (cosθ+sinθ)(cosθ-sinθ)=?
(a) a+ba-b
(b) a-ba+b
(c) b+ab-a
(d) b-ab+a

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)

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Question 13:

If tan θ = 17, then (cosec2θ-sec2θ)(cosec2θ+sec2θ)=? = ?
(a) -23
(b) -34
(c) 23
(d) 34

Answer:

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



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Question 14:

If tan θ=815 then cosec θ = ?
(a) 178

(b) 817

(c) 1517

(d) 1715

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).

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Question 15:

If 5tan θ=3 then 5sin θ-cos θ5sin θ+cos θ=?
(a) 23

(b) 13

(c) 12

(d) 35

Answer:


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

Hence, the correct answer is option (c).

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Question 16:

If in ΔABC, ∠C = 90°. Then, the value of cos(A + B) is

(a) 0

(b) 1

(c) 12

(d) 35

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).

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Question 17:

If cos A + cos2 A = 1, then (sin2 A + sin4 A) = ?
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

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

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Question 18:

If 3 cot θ = 4, then (5sinθ+3cosθ)(5sinθ-3cosθ) = ?
(a) 13
(b) 3
(c) 19
(d) 9

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]

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Question 19:

If 2x = sec A and 2x = tan A, then 2x2-1x2=? =?
(a) 12
(b) 14
(c) 18
(d) 116

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]

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Question 20:

If 3x = cosec θ and 3x = cot θ, than 3x2-1x2=?
(a) 127
(b) 181
(c) 13
(d) 19

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]

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Question 21:

If tan θ = 3, then sec θ = ?
(a) 23
(b) 32
(c) 12
(d) 2

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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