Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 8 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among Class 10 students for Math Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 8 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 332:

Question 1:

(i) (1 − cos2θ) cosec2θ = 1
(ii) (1 + cot2θ) sin2θ = 1

Answer:

(i) LHS=(1cos2θ)cosec2θ             =sin2θ cosec2θ         (cos2θ+sin2θ=1)             =1cosec2θ×cosec2θ             =1Hence, LHS = RHS(ii) LHS=(1+cot2θ)sin2θ             =cosec2θ sin2θ        (cosec2θcot2θ=1)             =1sin2θ×sin2θ             =1Hence, LHS=RHS

Page No 332:

Question 2:

(i) (sec2θ − 1) cot2θ = 1
(ii) (sec2θ − 1) (cosec2θ − 1) = 1
(iii) (1− cos2θ) sec2θ = tan2θ

Answer:

(i) LHS=(sec2θ1)cot2θ             =tan2θ×cot2θ       (sec2θtan2θ=1)             =1cot2θ×cot2θ             =1             =RHS(ii) LHS=(sec2θ1)(cosec2θ1)              =tan2θ ×cot2θ       (sec2θtan2θ=1 and cosec2θcot2θ=1)              =tan2θ×1tan2θ              =1              =RHS(iii) LHS=(1cos2θ)sec2θ               =sin2θ×sec2θ       (sin2θ+cos2θ=1)               =sin2θ×1cos2θ               =sin2θcos2θ              =tan2θ              =RHS

Page No 332:

Question 3:

(i) sin2θ+11+tan2θ=1
(ii) 11+tan2θ+11+cot2θ=1

Answer:

(i) LHS=sin2θ+1(1+tan2θ)              =sin2θ+1sec2θ      (sec2θtan2θ=1)              =sin2θ+cos2θ              =1              =RHS(ii) LHS=1(1+tan2θ)+1(1+cot2θ)              =1sec2θ+1cosec2θ              =cos2θ+sin2θ              =1              =RHS

Page No 332:

Question 4:

(i) (1 + cos θ) (1 − cos θ) (1 + cot2θ) = 1
(ii) cosec θ (1 + cos θ) (cosec θ − cot θ) = 1

Answer:

(i) LHS=(1+cosθ)(1cosθ)(1+cot2θ)             =(1cos2θ)cosec2θ             =sin2θ×cosec2θ             =sin2θ×1sin2θ             =1             =RHS(ii) LHS=cosecθ(1+cosθ)(cosecθcotθ)              =(cosecθ+cosecθ×cosθ)(cosecθcotθ)              =(cosecθ+1sinθ×cosθ)(cosecθcotθ)             =(cosecθ+cotθ)(cosecθcotθ)             =cosec2θcot2θ                          (cosec2θcot2θ=1)             =1             =RHS

Page No 332:

Question 5:

Prove each of the following identities:

i cot2θ-1sin2θ=-1ii tan2θ-1cos2θ=-1iii cos2θ+11+cot2θ=1

Answer:

iLHS=cot2θ-1sin2θ=cos2θsin2θ-1sin2θ=cos2θ-1sin2θ=-sin2θsin2θ=-1=RHS

iiLHS=tan2θ-1cos2θ=sin2θcos2θ-1cos2θ=sin2θ-1cos2θ=-cos2θcos2θ=-1=RHS

iiiLHS=cos2θ+11+cot2θ=cos2θ+1cosec2θ=cos2θ+sin2θ=1=RHS

Page No 332:

Question 6:

Prove that 11+sinθ+11-sinθ=2sec2θ

Answer:

LHS=11+sinθ+11-sinθ=1-sinθ+1+sinθ1+sinθ1-sinθ=21-sin2θ=2cos2θ=2sec2θ=RHS



Page No 333:

Question 7:

(i) sec θ (1 − sin θ) (sec θ + tan θ) = 1
(ii) sin θ(1 + tan θ) + cos θ(1 + cot θ) = (sec θ + cosec θ)

Answer:

(i) LHS=secθ(1sinθ)(secθ+tanθ)              =(secθsecθsinθ)(secθ+tanθ)              =(secθ1cosθ×sinθ)(secθ+tanθ)              =(secθtanθ)(secθ+tanθ)              =sec2θtan2θ              =1              =RHS     (ii) LHS=sinθ(1+tanθ)+cosθ(1+cotθ)              =sinθ+sinθ×sinθcosθ+cosθ+cosθ×cosθsinθ              =cosθsin2θ+sin3θ+cos2θsinθ+cos3θcosθsinθ              =(sin3θ+cos3θ)+(cosθsin2θ+cos2θsinθ)cosθsinθ              =(sinθ+cosθ)(sin2θsinθcosθ+cos2θ)+sinθcosθ(sinθ+cosθ)cosθsinθ              =(sinθ+cosθ)(sin2θ+cos2θsinθcosθ+sinθcosθ)cosθsinθ              =(sinθ+cosθ)(1)cosθsinθ              =sinθcosθsinθ+cosθcosθsinθ              =1cosθ+1sinθ              =secθ+cosecθ              =RHS

Page No 333:

Question 8:

(i) 1+cot2θ1+cosecθ=cosecθ
(ii) 1+tan2θ1+secθ=secθ

Answer:

(i) LHS=1+cot2θ(1+cosecθ)              =1+(cosec2θ1)(cosecθ+1)                (cosec2θ-cot2θ=1)              =1+(cosecθ+1)(cosecθ1)(cosecθ+1)              =1+(cosecθ1)              =cosecθ              =RHS(ii) LHS=1+tan2θ(1+secθ)              =1+(sec2θ1)(secθ+1)              =1+(secθ+1)(secθ1)(secθ+1)              =1+(secθ1)             =secθ             =RHS

Page No 333:

Question 9:

1+tan2θcotθcosec2θ=tanθ

Answer:

LHS=(1+tan2θ)cotθcosec2θ         =sec2θcotθcosec2θ         =1cos2θ×cosθsinθ1sin2θ         =1cosθsinθ×sin2θ         =sinθcosθ         =tanθ         =RHS
Hence, L.H.S. = R.H.S.

Page No 333:

Question 10:

tan2θ(1+tan2θ)+cot2θ(1+cot2θ)=1

Answer:

LHS=tan2θ(1+tan2θ)+cot2θ(1+cot2θ)         =tan2θsec2θ+cot2θcosec2θ             (sec2θtan2θ=1 and cosec2θcot2θ=1)         =sin2θcos2θ1cos2θ+cos2θsin2θ1sin2θ         =sin2θ+cos2θ        =1        =RHS

Hence, LHS = RHS

Page No 333:

Question 11:

sinθ1+cosθ+(1+cosθ)sinθ=2cosecθ

Answer:

LHS=sinθ(1+cosθ)+(1+cosθ)sinθ         =sin2θ+(1+cosθ)2(1+cosθ)sinθ         =sin2θ+1+cos2θ+2cosθ(1+cosθ)sinθ         =1+1+2cosθ(1+cosθ)sinθ         =2+2cosθ(1+cosθ)sinθ         =2(1+cosθ)(1+cosθ)sinθ         =2sinθ         =2cosecθ         =RHS

Hence, LHS = RHS

Page No 333:

Question 12:

tanθ(1-cotθ)+cotθ(1-tanθ)=(1+secθ cosecθ)

Answer:

LHS=tanθ(1cotθ)+cotθ(1tanθ)         =tanθ(1cosθsinθ)+cotθ(1sinθcosθ)         =sinθtanθ(sinθcosθ)+cosθcotθ(cosθsinθ)         =sinθ×sinθcosθcosθ×cosθsinθ(sinθcosθ)         =sin2θcosθcos2θsinθ(sinθcosθ)         =sin3θcos3θcosθsinθ(sinθcosθ)         =(sinθcosθ)(sin2θ+sinθcosθ+cos2θ)cosθsinθ(sinθcosθ)         =1+sinθcosθcosθsinθ         =1cosθsinθ+sinθcosθcosθsinθ         =secθcosecθ+1         =1+secθcosecθ         =RHS

Page No 333:

Question 13:

cos2θ(1-tanθ)+sin3θ(sinθ-cosθ)=(1+sinθ cosθ)

Answer:

cos2θ(1-tanθ)+sin3θ(sinθ-cosθ)=(1+sinθ cosθ)

LHS=cos2θ(1tanθ)+sin3θ(sinθcosθ)         =cos2θ(1sinθcosθ)+sin3θ(sinθcosθ)         =cos3θ(cosθsinθ)+sin3θ(sinθcosθ)         =cos3θsin3θ(cosθsinθ)         =(cosθsinθ)(cos2θ+cosθsinθ+sin2θ)(cosθsinθ)         =(sin2θ+cos2θ+cosθsinθ)         =(1+sinθcosθ)         =RHS

Hence, L.H.S. = R.H.S.

Page No 333:

Question 14:

cosθ(1-tanθ)+sin2θ(cosθ-sinθ)=(cosθ+sinθ)

Answer:

LHS=cosθ(1tanθ)sin2θ(cosθsinθ)         =cosθ(1sinθcosθ)sin2θ(cosθsinθ)        =cos2θ(cosθsinθ)sin2θ(cosθsinθ)        =cos2θsin2θ(cosθsinθ)        =(cosθ+sinθ)(cosθsinθ)(cosθsinθ)       =(cosθ+sinθ)       =RHS

Hence, LHS = RHS

Page No 333:

Question 15:

(1+tan2θ)(1+cot2θ)=1(sin2θ-sin4θ)

Answer:

LHS=(1+tan2θ)(1+cot2θ)         =sec2θ.cosec2θ             (sec2θtan2θ=1 and cosec2θcot2θ=1)         =1cos2θ.sin2θ         =1(1sin2θ)sin2θ         =1sin2θsin4θ        =RHS

Hence, LHS = RHS

Page No 333:

Question 16:

tanθ(1+tan2θ)2+cotθ(1+cot2θ)2=sinθ cosθ

Answer:

LHS=tanθ(1+tan2θ)2+cotθ(1+cot2θ)2         =tanθ(sec2θ)2+cotθ(cosec2θ)2         =tanθsec4θ+cotθcosec4θ         =sinθcosθ×cos4θ+cosθsinθ×sin4θ         =sinθcos3θ+cosθsin3θ         =sinθcosθ(cos2θ+sin2θ)         =sinθcosθ        =RHS

Hence, LHS = RHS

Page No 333:

Question 17:

(i) sin6θ+cos6θ=1-3sin2θcos2θ
(ii) sin2θ+cos4θ=cos2θ+sin4θ
(iii) cosec4θ-cosec2θ=cot4θ+cot2θ

Answer:

(i) LHS=sin6θ+cos6θ              =(sin2θ)3+(cos2θ)3              =(sin2θ+cos2θ)(sin4θsin2θcos2θ+cos4θ)              =1×{(sin2θ)2+2sin2θcos2θ+(cos2θ)23sin2θcos2θ}              =(sin2θ+cos2θ)23sin2θcos2θ              =(1)23sin2θcos2θ              =13sin2θcos2θ              =RHSHence, LHS = RHS(ii) LHS=sin2θ+cos4θ              =sin2θ+(cos2θ)2              =sin2θ+(1sin2θ)2              =sin2θ+12sin2θ+sin4θ              =1sin2θ+sin4θ              =cos2θ+sin4θ              =RHSHence, LHS = RHS(iii) LHS=cosec4θcosec2θ               =cosec2θ(cosec2θ1)               =cosec2θ×cot2θ             (cosec2θcot2θ=1)               =(1+cot2θ)×cot2θ               =cot2θ+cot4θ               =RHSHence, LHS = RHS

Page No 333:

Question 18:

(i) 1-tan2θ1+tan2θ=(cos2θ-sin2θ)
(ii) 1-tan2θcot2θ-1=tan2θ

Answer:

(i) LHS=1tan2θ1+tan2θ              =1sin2θcos2θ1+sin2θcos2θ              =cos2θsin2θcos2θ+sin2θ              =cos2θsin2θ1              =cos2θsin2θ              =RHS(ii) LHS=1tan2θcot2θ1               =1sin2θcos2θcos2θsin2θ1              =cos2θsin2θcos2θcos2θsin2θsin2θ              =sin2θcos2θ              =tan2θ              =RHS

Page No 333:

Question 19:

(i) tanθ(secθ-1)+tanθ(secθ+1)=2cosecθ
(ii) cotθ(cosecθ+1)+(cosecθ+1)cotθ=2secθ

Answer:

(i) LHS=tanθ(secθ1)+tanθ(secθ+1)             =tanθ{secθ+1+secθ1(secθ1)(secθ+1)}             =tanθ{2secθ(sec2θ1)}             =tanθ×2secθtan2θ             =2secθtanθ             =21cosθsinθcosθ             =21sinθ             =2cosecθ             =RHSHence, LHS = RHS(ii) LHS=cotθ(cosecθ+1)+(cosecθ+1)cotθ              =cot2θ+(cosecθ+1)2(cosecθ+1)cotθ              =cot2θ+cosec2θ+2cosecθ+1(cosecθ+1)cotθ             =cot2θ+cosec2θ+2cosecθ+cosec2θcot2θ(cosecθ+1)cotθ             =2cosec2θ+2cosecθ(cosecθ+1)cotθ             =2cosecθ(cosecθ+1)(cosecθ+1)cotθ             =2cosecθcotθ            =2×1sinθ×sinθcosθ            =2secθ            =RHSHence, LHS = RHS

Page No 333:

Question 20:

(i) secθ-1secθ+1=sin2θ(1+cosθ)2
(ii) secθ-tanθsecθ+tanθ=cos2θ(1+sinθ)2

Answer:

(i) LHS=secθ1secθ+1             =1cosθ11cosθ+1             =1cosθcosθ1+cosθcosθ            =1cosθ1+cosθ            =(1cosθ)(1+cosθ)(1+cosθ)(1+cosθ)           {Dividing the numerator and denominator by (1+cosθ)}            =1cos2θ(1+cosθ)2            =sin2θ(1+cosθ)2            =RHS(ii) LHS=secθtanθsecθ+tanθ              =1cosθsinθcosθ1cosθ+sinθcosθ              =1sinθcosθ1+sinθcosθ              =1sinθ1+sinθ              =(1sinθ)(1+sinθ)(1+sinθ)(1+sinθ)             {Dividing the numerator and denominator by (1+sinθ)}             =(1sin2θ)(1+sinθ)2              =cos2θ(1+sinθ)2              =RHS



Page No 334:

Question 21:

Prove each of the following identities:

i 1+sinθ1-sinθ=secθ+tanθii 1-cosθ1+cosθ=cosecθ-cotθiii 1+cosθ1-cosθ+1-cosθ1+cosθ=2cosecθ

Answer:

i LHS=1+sinθ1-sinθ=1+sinθ1-sinθ×1+sinθ1+sinθ=1+sinθ21-sin2θ=1+sinθ2cos2θ=1+sinθcosθ=1cosθ+sinθcosθ=secθ+tanθ=RHS

ii LHS=1-cosθ1+cosθ=1-cosθ1+cosθ×1-cosθ1-cosθ=1-cosθ21-cos2θ=1-cosθ2sin2θ=1-cosθsinθ=1sinθ-cosθsinθ=cosecθ-cotθ=RHS

(iii) LHS=1+cosθ1cosθ+1cosθ1+cosθ               =(1+cosθ)2(1cosθ)(1+cosθ)+(1cosθ)2(1+cosθ)(1cosθ)               =(1+cosθ)2(1cos2θ)+(1cosθ)2(1cos2θ)               =(1+cosθ)2sin2θ+(1cosθ)2sin2θ               =(1+cosθ)sinθ+(1cosθ)sinθ               =1+cosθ+1cosθsinθ               =2sinθ               =2cosecθ               =RHS

Page No 334:

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 334:

Question 23:

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 334:

Question 24:

(i) sinθ-cosθsinθ+cosθ+sinθ+cosθsinθ-cosθ=2(2sin2θ-1)
(ii) sinθ+cosθsinθ-cosθ+sinθ-cosθsinθ+cosθ=2(1-2cos2θ)

Answer:

(i) LHS=sinθcosθsinθ+cosθ+sinθ+cosθsinθcosθ              =(sinθcosθ)2+(sinθ+cosθ)2(sinθ+cosθ)(sinθcosθ)              =sin2θ+cos2θ2sinθcosθ+sin2θ+cos2θ+2sinθcosθsin2θcos2θ              =1+1sin2θ(1sin2θ)          (sin2θ+cos2θ=1)             =2sin2θ1+sin2θ             =22sin2θ1             =RHS(ii) LHS=sinθ+cosθsinθcosθ+sinθcosθsinθ+cosθ              =(sinθ+cosθ)2+(sinθcosθ)2(sinθcosθ)(sinθ+cosθ)              =sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ2sinθcosθ(sin2θcos2θ)              =1+1(1cos2θ)cos2θ       (sin2θ+cos2θ=1)              =212cos2θ              =RHS

Page No 334:

Question 25:

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 334:

Question 26:

(i) cosecθ+cotθcosecθ-cotθ=(cosecθ+cotθ)2=1+2cot2+2cosecθ cotθ
(ii) secθ+tanθsecθ-tanθ=(secθ+tanθ)2=1+2tan2θ+2secθ tanθ

Answer:

(i) Here, cosecθ+cotθcosecθcotθ=(cosecθ+cotθ)(cosecθ+cotθ)(cosecθcotθ)(cosecθ+cotθ)=(cosecθ+cotθ)2(cosec2θcot2θ)=(cosecθ+cotθ)21=(cosecθ+cotθ)2Again, (cosecθ+cotθ)2        =cosec2θ+cot2θ+2cosecθcotθ        =1+cot2θ+cot2θ+2cosecθcotθ        (cosec2θcot2θ=1)        =1+2cot2θ+2cosecθcotθ(ii) Here,secθ+tanθsecθtanθ             =(secθ+tanθ)(secθ+tanθ)(secθtanθ)(secθ+tanθ)             =(secθ+tanθ)2sec2θtan2θ             =(secθ+tanθ)21             =(secθ+tanθ)2Again, (secθ+tanθ)2         =sec2θ+tan2θ+2secθtanθ         =1+tan2θ+tan2θ+2secθtanθ         =1+2tan2θ+2secθtanθ

Page No 334:

Question 27:

(i) 1+cosθ+sinθ1+cosθ-sinθ=1+sinθcosθ
(ii) sinθ+1-cosθcosθ-1+sinθ=1+sinθcosθ

Answer:

(i) LHS=1+cosθ+sinθ1+cosθsinθ             ={(1+cosθ)+sinθ}{(1+cosθ)+sinθ}{(1+cosθ)sinθ}{(1+cosθ)+sinθ}          {Multiplying the numerator and denominator by (1+cosθ+sinθ)}             ={(1+cosθ)+sinθ}2{(1+cosθ)2sin2θ}             =1+cos2θ+2cosθ+sin2θ+2sinθ(1+cosθ)1+cos2θ+2cosθsin2θ             =2+2cosθ+2sinθ(1+cosθ)1+cos2θ+2cosθ(1cos2θ)             =2(1+cosθ)+2sinθ(1+cosθ)2cos2θ+2cosθ             =2(1+cosθ)(1+sinθ)2cosθ(1+cosθ)             =1+sinθcosθ            =RHS(ii)LHS=sinθ+1cosθcosθ1+sinθ              =(sinθ+1cosθ)(sinθ+cosθ+1)(cosθ1+sinθ)(sinθ+cosθ+1)       {Multiplying and dividing by 1+cosθ+sinθ }             =(sinθ+1)2cos2θ(sinθ+cosθ)212             =sin2θ+1+2sinθcos2θsin2θ+cos2θ+2sinθcosθ1            =sin2θ+sin2θ+cos2θ+2sinθcos2θ2sinθcosθ            =2sin2θ+2sinθ2sinθcosθ            =2sinθ(1+sinθ)2sinθcosθ            =1+sinθcosθ            =RHS

Page No 334:

Question 28:

sinθ(secθ+tanθ-1)+cosθ(cosecθ+cotθ-1)=1

Answer:

LHS=sinθ(secθ+tanθ1)+cosθ(cosecθ+cotθ1)        =sinθcosθ1+sinθcosθ+cosθsinθ1+cosθsinθ        =sinθcosθ[11+(sinθcosθ)+11(sinθcosθ)]        =sinθcosθ[1(sinθcosθ)+1+(sinθcosθ){1+(sinθcosθ)}{1(sinθcosθ)}]        =sinθcosθ[1sinθ+cosθ+1+sinθcosθ1(sinθcosθ)2]        =2sinθcosθ1(sin2θ+cos2θ2sinθcosθ)         =2sinθcosθ2sinθcosθ         =1         =RHS
Hence, LHS = RHS

Page No 334:

Question 29:

sinθ+cosθsinθ-cosθ+sinθ-cosθsinθ+cosθ=2(sin2θ-cos2θ)=2(2sin2θ-1)

Answer:

We havesinθ+cosθsinθcosθ+sinθcosθsinθ+cosθ         =(sinθ+cosθ)2+(sinθcosθ)2(sinθcosθ)(sinθ+cosθ)         =sin2θ+cos2θ+2sinθcosθ+sin2θ+cos2θ2sinθcosθsin2θcos2θ         =1+1sin2θcos2θ         =2sin2θcos2θAgain,2sin2θcos2θ         =2sin2θ(1sin2θ)         =22sin2θ1

Page No 334:

Question 30:

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 334:

Question 31:

(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 334:

Question 32:

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 334:

Question 33:

1(sec2θ-cos2θ)+1(cosec2θ-sin2θ)(sin2θ cos2θ)=1-sin2θcos2θ2+sin2θ cos2θ

Answer:

LHS={1sec2θcos2θ+1cosec2θsin2θ}(sin2θcos2θ)         ={cos2θ1cos4θ+sin2θ1sin4θ}(sin2θcos2θ)         ={cos2θ(1cos2θ)(1+cos2θ)+sin2θ(1sin2θ)(1+sin2θ)}(sin2θcos2θ)         =[cot2θ1+cos2θ+tan2θ1+sin2θ]sin2θcos2θ         =cos4θ1+cos2θ+sin4θ1+sin2θ         =(cos2θ)21+cos2θ+(sin2θ)21+sin2θ         =(1sin2θ)1+cos2θ+(1cos2θ)21+sin2θ        =(1sin2θ)2(1+sin2)+(1cos2θ)2(1+cos2θ)(1+sin2θ)(1+cos2θ)        =cos4θ(1+sin2θ)+sin4θ(1+cos2θ)1+sin2θ+cos2θ+sin2θcos2θ=cos4θ+cos4θsin2θ+sin4θ+sin4θcos2θ1+1+sin2θcos2θ=cos4θ+sin4θ+sin2θcos2θ(sin2θ+cos2θ)2+sin2θcos2θ=(cos2θ)2+(sin2θ)2+sin2θcos2θ(1)2+sin2θcos2θ=(cos2θ+sin2θ)2-2sin2θcos2θ+sin2θcos2θ(1)2+sin2θcos2θ       =12+cos2θsin2θ2cos2θsin2θ2+sin2θcos2θ        =1cos2θsin2θ2+sin2θcos2θ        =RHS

Page No 334:

Question 34:

(sin A-sin B)(cos A+cos B)+(cos A-cos B)(sin A+sin B)=0

Answer:

LHS=(sinAsinB)(cosA+cosB)+(cosAcosB)(sinA+sinB)          =(sinAsinB)(sinA+sinB)+(cosAcosB)(cosAcosB)(cosA+cosB)(sinA+sinB)          =sin2Asin2B+cos2Acos2B(cosA+cosB)(sinA+sinB)         =0(cosA+cosB)(sinA+sinB)         =0         =RHS



Page No 335:

Question 35:

tan A+tan Bcot A+cot B=tan A tan B

Answer:

LHS=tanA+tanBcotA+cotB         =tanA+tanB1tanA+1tanB         =tanA+tanBtanA+tanBtanAtanB         =tanAtanB(tanA+tanB)(tanA+tanB)         =tanAtanB         =RHS

Hence, LHS = RHS

Page No 335:

Question 36:

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 335:

Question 37:

Prove that sinθ-2sin3θ=2cos3θ-cosθtanθ

Answer:

RHS=2cos3θ-cosθtanθ=2cos2θ-1cosθ×sinθcosθ=21-sin2θ-1sinθ=2-2sin2θ-1sinθ=1-2sin2θsinθ=sinθ-2sin3θ=LHS



Page No 343:

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 343:

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 343:

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 343:

Question 4:

If (sec θ + tan θ) = m and (sec θ − tan θ) = n, show that mn = 1.

Answer:

We have (secθ+tanθ)=m        ...(i)Again, (secθtanθ)=n         ...(ii)Now, multiplying (i) and (ii), we get:(secθ+tanθ)×(secθtanθ)=mn=>sec2θtan2θ=mn=>1=mn       [sec2θtan2θ=1]mn=1

Page No 343:

Question 5:

If (cosec θ + cot θ) = m and (cosec θ − cot θ) = n, show that mn = 1.

Answer:

We have (cosecθ+cotθ)=m      ...(i)Again, (cosecθ-cotθ)=n         ...(ii)Now, multiplying (i) and (ii), we get:(cosecθ+cotθ)×(cosecθcotθ)=mn=>cosec2θcot2θ=mn=>1=mn       [cosec2θcot2θ=1]mn=1

Page No 343:

Question 6:

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 343:

Question 7:

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 343:

Question 8:

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 343:

Question 9:

If (cosec θ − sin θ) = a3and (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 343:

Question 10:

If (2 sin θ + 3 cos θ) = 2, show that (3 sin θ − 2 cos θ) = ± 3.

Answer:

Given, (2sinθ+3cosθ)=2      ...(i)We have (2sinθ+3cosθ)2+(3sinθ2cosθ)2=4sin2θ+9cos2θ+12sinθcosθ+9sin2θ+4cos2θ12sinθcosθ  =4(sin2θ+cos2θ)+9(sin2θ+cos2θ)  =4+9=13i.e., (2sinθ+3cosθ)2+(3sinθ2cosθ)2=13=>22+(3sinθ2cosθ)2=13=>(3sinθ2cosθ)2=134=>(3sinθ2cosθ)2=9=>(3sinθ2cosθ)=±3

Page No 343:

Question 11:

If sinθ+cosθ=2cosθ, show that cotθ=2+1.

Answer:

We have,sinθ+cosθ=2cosθDividing both sides by sinθ, we getsinθsinθ+cosθsinθ=2cosθsinθ1+cotθ=2cotθ2cotθ-cotθ=12-1cotθ=1
cotθ=12-1cotθ=12-1×2+12+1cotθ=2+12-1cotθ=2+11 cotθ=2+1

Page No 343:

Question 12:

If cos θ + sin θ = 2 sin θ, show that sin θ − cos θ = 2 cos θ.

Answer:

Given: cosθ+sinθ=2sinθWe have (sinθ+cosθ)2+(sinθcosθ)2=2(sin2θ+cos2θ)=>(2sinθ)2+(sinθcosθ)2=2=>2sin2θ+(sinθcosθ)2=2=>(sinθcosθ)2=22sin2θ=>(sinθcosθ)2=2(1sin2θ)=>(sinθcosθ)2=2cos2θ=>(sinθcosθ)=2cosθHence proved.

Page No 343:

Question 13:

If secθ+tanθ=p, prove that

i secθ=12p+1p       ii tanθ=12p-1p       iii sinθ=p2-1p2+1

Answer:

iWe have,secθ+tanθ=p         .....1secθ+tanθ1×secθ-tanθsecθ-tanθ=psec2θ-tan2θsecθ-tanθ=p1secθ-tanθ=psecθ-tanθ=1p     .....2Adding 1 and 2, we get2secθ=p+1psecθ=12p+1p

ii Subtracting 2 from 1, we get2tanθ=p-1ptanθ=12p-1p

iii Using i and ii, we getsinθ=tanθsecθ=12p-1p12p+1p=p2-1pp2+1p sinθ=p2-1p2+1

Page No 343:

Question 14:

If tan A = n tan B and sin A = m sin B, prove that cos2A = (m2-1)n2-1.

Answer:

We have tanA=ntanB => cotB=ntanA          ...(i)Again, sinA=msinB=>cosecB=msinA         ...(ii)Squaring (i) and (ii) and subtracting (ii) from (i), we getm2sin2An2tan2A=cosec2Bcot2B=> m2sin2An2cossin2A=1=>m2n2cos2A=sin2A=>m2n2cos2A=1cos2A=>n2cos2Acos2A=m21=>cos2A(n21)=(m21)=>cos2A=(m21)(n21)∴ cos2A=(m21)(n21)

Page No 343:

Question 15:

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 345:

Question 1:

Write the value of 1-sin2θsec2θ.

Answer:

1-sin2θsec2θ=cos2θ×1cos2θ=1

Page No 345:

Question 2:

Write the value of 1-cos2θcosec2θ.

Answer:

1-cos2θcosec2θ=sin2θ×1sin2θ=1

Page No 345:

Question 3:

Write the value of 1+tan2θcos2θ.

Answer:

1+tan2θcos2θ=sec2θ×1sec2θ=1

Page No 345:

Question 4:

Write the value of 1+cot2θsin2θ.

Answer:

1+cot2θsin2θ=cosec2θ×1cosec2θ=1

Page No 345:

Question 5:

Write the value of sin2θ+11+tan2θ.

Answer:

sin2θ+11+tan2θ=sin2θ+1sec2θ=sin2θ+cos2θ=1

Page No 345:

Question 6:

Write the value of cot2θ-1sin2θ.

Answer:

cot2θ-1sin2θ=cot2θ-cosec2θ=-1

Page No 345:

Question 7:

Write the value of sinθ cos90°-θ+cosθ sin90°-θ.

Answer:

sinθ cos90°-θ+cosθ sin90°-θ=sinθ sinθ+cosθ cosθ=sin2θ+cos2θ=1

Page No 345:

Question 8:

Write the value of cosec290°-θ-tan2θ.

Answer:

cosec290°-θ-tan2θ=sec2θ-tan2θ=1

Page No 345:

Question 9:

Write the value of sec2θ1+sinθ1-sinθ.                   [CBSE 2009]

Answer:

sec2θ1+sinθ1-sinθ=sec2θ1-sin2θ=1cos2θ×cos2θ=1

Page No 345:

Question 10:

Write the value of cosec2θ1+cosθ1-cosθ.

Answer:

cosec2θ1+cosθ1-cosθ=cosec2θ1-cos2θ=1sin2θ×sin2θ=1

Disclaimer: The question given in the textbook is incorrect. There should be cosθ instead sinθ. The solution provided here is of the same.

Page No 345:

Question 11:

Write the value of sin2θ cos2θ1+tan2θ1+cot2θ.

Answer:

sin2θ cos2θ1+tan2θ1+cot2θ=sin2θ cos2θ sec2θ cosec2θ=sin2θ×cos2θ×1cos2θ×1sin2θ=1

Page No 345:

Question 12:

Write the value of 1+tan2θ1+sinθ1-sinθ.                [CBSE 2008]

Answer:

1+tan2θ1+sinθ1-sinθ=sec2θ1-sin2θ=1cos2θ×cos2θ=1



Page No 346:

Question 13:

Write the value of 3cot2θ-3cosec2θ.

Answer:

3cot2θ-3cosec2θ=3cot2θ-cosec2θ=3-1=-3

Page No 346:

Question 14:

Write the value of 4tan2θ-4cos2θ.

Answer:

4tan2θ-4cos2θ=4tan2θ-4sec2θ=4tan2θ-sec2θ=4-1=-4

Page No 346:

Question 15:

Write the value of tan2θ-sec2θcot2θ-cosec2θ.

Answer:

tan2θ-sec2θcot2θ-cosec2θ=-1-1=1

Page No 346:

Question 16:

If sinθ=12, then write the value of 3cot2θ+3.                     CBSE 2009

Answer:

As, sinθ=12So, cosecθ=1sinθ=2         .....i

Now,

3cot2θ+3=3cot2θ+1=3cosec2θ=322                   Using i=34=12

Page No 346:

Question 17:

If cosθ=23, the write the value of 4+4tan2θ.

Answer:

4+4tan2θ=41+tan2θ=4sec2θ=4cos2θ=4232=449=4×94=9

Page No 346:

Question 18:

If cosθ=725, then write the value of tanθ+cotθ.           CBSE 2008

Answer:

As sin2θ=1-cos2θ=1-7252=1-49625=625-49625sin2θ=576625sinθ=576625sinθ=2425

Now,tanθ+cotθ=sinθcosθ+cosθsinθ=sin2θ+cos2θcosθ sinθ=1725×2425=1168625=625168

Page No 346:

Question 19:

If cosθ=23, then write the value of secθ-1secθ+1.

Answer:

secθ-1secθ+1=1cosθ-111cosθ+11=1-cosθcosθ1+cosθcosθ=1-cosθ1+cosθ=11-2311+23=1353=15

Page No 346:

Question 20:

If 5tanθ=4, then write the value of cosθ-sinθcosθ+sinθ.

Answer:

We have,5tanθ=4tanθ=45Now,cosθ-sinθcosθ+sinθ=cosθcosθ-sinθcosθcosθcosθ+sinθcosθ                   Dividing numerator and denominator by cosθ=1-tanθ1+tanθ=11-4511+45=1595=19

Page No 346:

Question 21:

If 3cotθ=4, then write the value of 2cosθ+sinθ4cosθ-sinθ.

Answer:

We have,3cotθ=4cotθ=43Now,2cosθ+sinθ4cosθ-sinθ=2cosθsinθ+sinθsinθ4cosθsinθ-sinθsinθ              Dividing numerator and denominator by sinθ=2cotθ+14cotθ-1=2×43+14×43-1=83+11163-11=8+3316-33=113133=1113

Page No 346:

Question 22:

If cotθ=13, then write the value of 1-cos2θ2-sin2θ.

Answer:

We have,cotθ=13cotθ=cotπ3θ=π3Now,1-cos2θ2-sin2θ=1-cos2π32-sin2π3=1-1222-322=11-1421-34=3454=35

Page No 346:

Question 23:

If tanθ=15, then write the value of cosec2θ-sec2θcosec2θ+sec2θ.

Answer:

cosec2θ-sec2θcosec2θ+sec2θ=1+cot2θ-1+tan2θ1+cot2θ+1+tan2θ=1+1tan2θ-1+tan2θ1+1tan2θ+1+tan2θ=1+1tan2θ-1-tan2θ1+1tan2θ+1+tan2θ=1tan2θ-tan2θ1tan2θ+tan2θ+2=512-152512+152+2=51-1551+15+21=245365=2436=23

Page No 346:

Question 24:

If cotA=43 and A+B=90°, then what is the value of tanB?

Answer:

We have,cotA=43cot90°-B=43             As, A+B=90° tanB=43

Page No 346:

Question 25:

If cosB=35 and A+B=90°, then find the value of sinA.

Answer:

We have,cosB=35cos90°-A=35             As, A+B=90° sinA=35

Page No 346:

Question 26:

If 3sinθ=cosθ and θ is an acute angle, then find the value of θ.

Answer:

We have,3sinθ=cosθsinθcosθ=13tanθ=13tanθ=tan30° θ=30°

Page No 346:

Question 27:

Write the value of tan10° tan20° tan70° tan80°.

Answer:

tan10° tan20° tan70° tan80°=cot90°-10° cot90°-20° tan70° tan80°=cot80° cot70° tan70° tan80°=1tan80°×1tan70°×tan70°×tan80°=1

Page No 346:

Question 28:

Write the value of tan1° tan2° ... tan89°.

Answer:

tan1° tan2° ... tan89°=tan1° tan2° tan3° ... tan45° ...tan87° tan88° tan89°=tan1° tan2° tan3° ... tan45° ...cot90°-87° cot90°-88° cot90°-89°=tan1° tan2° tan3° ... tan45° ...cot3° cot2° cot1°=tan1°×tan2°×tan3°×...×1×...×1tan3°×1tan2°×1tan1°=1

Page No 346:

Question 29:

Write the value of cos1° cos2° ... cos180°.

Answer:

cos1° cos2° ... cos180°=cos1° cos2° ... cos90° ... cos180°=cos1° cos2° ... 0 ... cos180°=0

Page No 346:

Question 30:

If tanA=512, then find the value of sinA+cosAsecA.                CBSE 2008

Answer:

sinA+cosAsecA=sinA+cosA1cosA=sinAcosA+cosAcosA=tanA+1=512+11=5+1212=1712

Page No 346:

Question 31:

If sinθ=cosθ-45°, where θ is acute, then find the value of θ.

Answer:

We have,sinθ=cosθ-45°cos90°-θ=cosθ-45°Comparing both sides, we get90°-θ=θ-45°θ+θ=90°+45°2θ=135°θ=1352° θ=67.5°

Page No 346:

Question 32:

Find the value of sin50°cos40°+cosec40°sec50°-4cos50° cosec40°.

Answer:

sin50°cos40°+cosec40°sec50°-4cos50° cosec40°=cos90°-50°cos40°+sec90°-40°sec50°-4sin90°-50° cosec40°=cos40°cos40°+sec50°sec50°-4sin40°×1sin40°=1+1-4=-2

Page No 346:

Question 33:

Find the value of sin48° sec42°+cos48° cosec42°.

Answer:

sin48° sec42°+cos48° cosec42°=sin48° cosec90°-42°+cos48° sec90°-42°=sin48° cosec48°+cos48° sec48°=sin48°×1sin48°+cos48°×1cos48°=1+1=2

Page No 346:

Question 34:

If x=asinθ and y=bcosθ, then write the value of b2x2+a2y2.

Answer:

b2x2+a2y2=b2asinθ2+a2bcosθ2=b2a2sin2θ+a2b2cos2θ=a2b2sin2θ+cos2θ=a2b21=a2b2

Page No 346:

Question 35:

If 5x=secθ and 5x=tanθ, then find the value of 5x2-1x2.                  CBSE 2010

Answer:

5x2-1x2=255x2-1x2=1525x2-25x2=155x2-5x2=15secθ2-tanθ2=15sec2θ-tan2θ=151=15

Page No 346:

Question 36:

If cosecθ=2x and cotθ=2x, then find the value of 2x2-1x2.                   CBSE 2010

Answer:

2x2-1x2=42x2-1x2=124x2-4x2=122x2-2x2=12cosecθ2-secθ2=12cosec2θ-sec2θ=121=12



Page No 347:

Question 37:

If secθ+tanθ=x, then find the value of secθ.

Answer:

We have,secθ+tanθ=x                   .....isecθ+tanθ1×secθ-tanθsecθ-tanθ=xsec2θ-tan2θsecθ-tanθ=x1secθ-tanθ=x1secθ-tanθ=1x               .....iiAdding i and ii, we get2secθ=x+1x2secθ=x2+1x secθ=x2+12x

Page No 347:

Question 38:

Find the value of cos38° cosec52°tan18° tan35° tan60° tan72° tan55°.

Answer:

cos38° cosec52°tan18° tan35° tan60° tan72° tan55°=cos38° sec90°-52°cot90°-18° cot90°-35° tan60° tan72° tan55°=cos38° sec38°cot72° cot55° tan60° tan72° tan55°=cos38°×1cos38°1tan72°×1tan55°×3×tan72°×tan55°=13

Page No 347:

Question 39:

If sinθ=x, then write the value of cotθ.

Answer:

cotθ=cosθsinθ=1-sin2θsinθ=1-x22

Page No 347:

Question 40:

If secθ=x, then write the value of tanθ.

Answer:

As, tan2θ=sec2θ-1So, tanθ=sec2θ-1=x2-1



Page No 350:

Question 1:

Choose the correct answer of the following question:sec30°cosec60°=?a 23          b 32          c 3          d1

Answer:

sec30°cosec60°=sec30°sec90°-60°=sec30°sec30°=1Hence, the correct answer is option d.

Page No 350:

Question 2:

tan35°cot55°+cot78°tan12°=?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

(c) 2We have: tan350cot550+cot780tan12o=tan350cot(900350)+cot(900120)tan120=tan350tan350+tan120tan120         [cot(900θ)=tanθ]=1+1=2

Page No 350:

Question 3:

tan 10° tan 15° tan 75° tan 80° = ?
(a) 3
(b) 13
(c) −1
(d) 1

Answer:

(d) 1We have:   tan100tan150tan750tan800=tan100×tan150×tan(900150)×tan(900100)=tan100×tan150×cot150×cot100                         [tan(900θ)=cotθ]=1

Page No 350:

Question 4:

tan 5° tan 25° tan 30° tan 65° tan 85° = ?
(a) 3
(b) 13
(c) 1
(d) none of these

Answer:

The correct option is (b).We have:tan50tan250tan300.tan650tan850= tan50tan250tan300tan(900250) tan(90050)=tan50tan250×13×cot250cot50      [tan(900θ)=cotθ and tan300=13]=13

Page No 350:

Question 5:

cos 1° cos 2° cos 3° ... cos 180° = ?
(a) −1
(b) 1
(c) 0
(d) 12

Answer:

(c) 0cos10cos20cos30...cos1800=cos10cos20cos30...cos 900...cos(180)0=0     [cos 90° =0]

Page No 350:

Question 6:

2sin263°+1+2sin227°3cos217°-2+3cos273°=?
(a) 32
(b) 23
(c) 2
(d) 3

Answer:

(d)3  Given: 2sin2630+1+2sin22703cos21702+3cos2730=2(sin2630+sin2270)+13(cos2170+cos2730)2=2[sin2630+sin2(900630)]+13[cos2170+cos2(900170)]2=2(sin2630+cos2630)+13(cos2170+sin2170)2        [sin(900θ)=cosθ and cos(900θ)=sinθ]=2×1+13×12                     [sin2θ+cos2θ=1]=2+132=31=3

Page No 350:

Question 7:

(sin 43°cos 47° + cos 43°sin 47°) = ?
(a) sin 4°
(b) cos 4°
(c) 1
(d) 0

Answer:

(c) 1We have:    (sin430cos470+cos430sin470)=sin430cos(900430)+cos430sin(900430)=sin430sin430+cos430cos430              [cos(900θ)=sinθ and sin(900θ)=cosθ]=sin2430+cos2430=1

Page No 350:

Question 8:

Choose the correct answer of the following question:sec70° sin20°+cos20° cosec70°=?a 0               b 1               c -1               d 2

Answer:

sec70° sin20°+cos20° cosec70°=cosec90°-70° sin20°+cos20° sec90°-70°=cosec20° sin20°+cos20° sec20°=1sin20°×sin20°+cos20°×1cos20°=1+1=2Hence, the correct answer is option d.

Page No 350:

Question 9:

If sin 3A = cos (A − 10°), where 3A is an acute angle, then ∠A = ?
(a) 35°
(b) 25°
(c) 20°
(d) 45°

Answer:

(b) 25°We have:      [sin3A=cos(A100)]=>cos(9003A)=cos(A100)         [sinθ=cos(900θ)]=>9003A=A100=>4A=100=>A=1002541=>A=250



Page No 351:

Question 10:

Choose the correct answer of the following question:If sec4A=cosecA-10° and 4A is acute, then A=?a 20°              b 30°               c 40°                 d 50°

Answer:

We have,sec4A=cosecA-10°cosec90°-4A=cosecA-10°Comparing both sides, we get90°-4A=A-10°4A+A=90°+10°5A=100°A=100°5 A=20°Hence, the correct answer is option a.

Page No 351:

Question 12:

If cos(α + β) = 0, then sin(α − β) = ?
(a) sin α
(b) cos β
(c) sin 2α
(d) cos 2β

Answer:

(d) cos 2β

We have:        cos(α+β)=0=>cos(α+β)=cos900=>α+β=900=>α=900β         ...(i)Now, sin(αβ)=sin[(900β)β]                       [Using (i)]=sin(9002β)=cos2β                                 [sin(900θ)=cosθ]

Page No 351:

Question 13:

sin (45° + θ) − cos (45° − θ) = ?
(a) 2 sin θ
(b) 2 cos θ
(c) 0
(d) 1

Answer:

(c) 0We have:     [sin(450+θ)cos(450θ)]=[sin{900(450θ)}cos(450θ)]=[cos(450θ)cos(450θ)]           [sin(900θ)=cosθ]=0

Page No 351:

Question 14:

(sin 79° cos 11° + cos 79° sin 11°) = ?
(a) 12
(b) 12
(c) 0
(d) 1

Answer:

(d) 1We have: (sin790cos110+cos790sin110)=sin790cos(900790)+cos790sin(900790)=sin790sin790cos790cos790             [cos(900θ)=sinθ and sin(900θ)=cosθ ]=sin2790+cos2790=1

Page No 351:

Question 15:

(cosec257° − tan233°) = ?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

  (b) 1We have:(cosec2570tan2330)=[cosec2(900330)tan2330]=(sec2330tan2330)                  [cosec(900θ)=secθ]=1                                                 [sec2θtan2θ=1]

Page No 351:

Question 16:

2tan230°sec252°sin238°(cosec270°-tan220°)=?
(a) 32
(b) 23
(c) 2
(d) 12

Answer:

(b) 23

We have: [2tan2300sec2520sin2380cosec2700tan2200]=[2×(13)2sec2520{sin2(900520)}{cosec2(900200)}tan2200]=[23×sec2520.cos2520sec2200tan2200]           [sin(900θ)=cosθ and cosec(900θ)=secθ]=23×11                               [sec2θtan2θ=1]=23

Page No 351:

Question 17:

(sin222°+sin268°)(cos222°+cos268°)+sin263°+cos63°sin27=?
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

(c) 2  We have:  [sin2220+sin2680cos2220+cos268+sin2630+cos630sin270]=[sin2220+sin2(900220)cos2(900680)+cos2680+sin2630+cos630{sin(900630)}]=[sin2220+cos2220sin2680+cos2680+sin2630+cos630cos630]      [sin(900θ)=cosθ and cos(900θ)=sinθ]=[11+sin2630+cos2630]                     [sin2θ+cos2θ=1]=1+1=2

Page No 351:

Question 18:

cot(90°-θ)·sin(90°-θ)sinθ+cot40°tan50°-(cos220°+cos270°)=?
(a) 3
(b) 2
(c) 1
(d) 0

Answer:

(c) 1We have:  [cot(900θ).sin(900θ)sinθ+cot400tan500(cos2200+cos2700)]=[tanθ.cosθsinθ+cot(900500)tan500{cos2(900700)+cos2700}]          [cot(900θ)=tanθ and sin(900θ)=cosθ]=[sinθcosθ.cosθsinθ+tan500tan500(sin2700+cos2700)]      [cos(900θ)=sinθ ]=(sinθsinθ+11)=1+11=1

Page No 351:

Question 19:

cos38° cosec52°tan18° tan35° tan60° tan72° tan55°=?
(a) 3
(b) 13
(c) 13
(d) 23

Answer:

(c) 13

We have:      [cos380cosec520tan180tan350tan600tan720tan550]=[cos380cosec(900380)tan180tan350×3×tan(900180)tan(900350)]        [cosec(900θ)=secθ and tan(900θ)=cotθ]=[cos380sec380tan180tan350×3×cot180cot350]=[1sec380×sec3801cot1801cot350×3cot180cot350]=13

Page No 351:

Question 20:

If 2 sin 2θ = 3, then θ = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

(a) 30o

    2 sin 2θ=3sin 2θ=32=sin 60osin 2θ=sin 60o2θ=60oθ=30o

Page No 351:

Question 21:

If 2cos 3θ = 1, then θ = ?
(a) 10°
(b) 15°
(c) 20°
(d) 30°

Answer:

(c) 20o

     2 cos 3θ=1 cos 3θ=12cos 3θ=cos 60o                    cos 60o=12 3θ=60oθ=60o3=20o

Page No 351:

Question 22:

If 3 tan 2θ − 3 = 0, then θ = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

Answer:

(b) 30o

3tan 2θ -3=03tan 2θ =3tan 2θ=33tan 2θ=3         [tan 60o =3 ]tan 2θ=tan 60o2θ=60oθ=30o 

Page No 351:

Question 23:

If tan x = 3 cot x, then x = ?
(a) 45°
(b) 60°
(c) 30°
(d) 15°

Answer:

(b) 60o

     tan x=3 cot xtan xcot x=3 tan2 x=3               cot x=1tan x tan x=3=tan 60o x= 60o



Page No 352:

Question 24:

If x tan 45° cos 60° = sin 60° cot 60°, then x = ?
(a) 1
(b) 12
(c) 12
(d) 3

Answer:

(a) 1

     x tan 45o cos 60o = sin 60o cot 60ox 1 12=3213x 12=12x=1

Page No 352:

Question 25:

If (tan245° − cos230°) = x sin 45° cos 45°, then x = ?
(a) 2
(b) −2
(c) 12
(d) -12

Answer:

(c) 12

tan2 45o - cos2 30o = x sin 45o cos 45ox=tan2 45o-cos2 30o sin 45o cos 45o=12-32212×12=1-3412=1412=14×2=12

Page No 352:

Question 26:

(sec260°− 1) = ?
(a) 2
(b) 3
(c) 4
(d) 0

Answer:

(b) 3

sec260o − 1 = (2)2 − 1 = 4 − 1 = 3

Page No 352:

Question 27:

(cos 0° + sin 30° + sin 45°) (sin 90° + cos 60° − cos 45°) = ?
(a) 56
(b) 58
(c) 35
(d) 74

Answer:

(d) 74

cos 0o+sin 30o+sin 45osin 90o+cos 60o-cos 45o= 1+12+121+12-12=32+1232-12=322-122=94 -12=9-24=74

Page No 352:

Question 28:

(sin230° + 4cot245° − sec260°) = ?
(a) 0
(b) 14
(c) 4
(d) 1

Answer:

(b) 14

sin2 30o+4 cot2 45o-sec2 60o=122+4×12-22=14+4-4=14

Page No 352:

Question 29:

(3cos260° + 2cot230° − 5sin245°) = ?
(a) 136
(b) 174
(c) 1
(d) 4

Answer:

(b) 174

3 cos2 60o+2 cot2 30o-5 sin2 45o = 3×122+2×32-5×122 = 34+6-52=3+24-104=174

Page No 352:

Question 30:

cos230°cos245°+4sec260°+12cos290°-2tan260°=?
(a) 738
(b) 758
(c) 818
(d) 838

Answer:

(d) 838

cos2 30o cos2 45o+4 sec2 60o+12 cos2 90o-2 tan2 60o= 322×122+4×22+12×02-2×32= 34×12+16-6=38+10=3+808=838

Page No 352:

Question 31:

If cosec θ = 10, then sec θ = ?
(a) 310
(b) 103
(c) 110
(d) 210

Answer:

(b) 103
Let us first draw a right ABC right angled at B and A=θ.
Given: cosec θ10, but sin θ1cosec θ = 110
Also, sin θ = PerpendicularHypotenuse = BCAC
So, BCAC = 110
Thus, BC = k and AC = 10k


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2-BC2
AB2= (10k)2 - (k)2
AB2 = 9k2
AB = 3k
∴ sec θ = ACAB = 10k3k=103

Page No 352:

Question 32:

If tan θ = 815, then cosec θ = ?
(a) 178
(b) 817
(c) 1715
(d) 1517

Answer:

(a) 178

Let us first draw a right ABC right angled at B and A=θ.
Given: tan θ = 815, but tan θ = BCAB
So, BCAB = 815
Thus, BC = 8k and AB = 15k


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AC2= (15k)2 + (8k)2
⇒ AC2 = 289k2
⇒ AC = 17k
∴ cosec θ = ACBC=17k8k=178

Page No 352:

Question 33:

If sin θ ab, then cos θ = ?
(a) bb2-a2
(b) b2-a2b
(c) ab2-a2
(d) ba

Answer:

(b) b2-a2b
Let us first draw a right ABC right angled at B and A=θ.
Given: sin θ = ab, but sin θ = BCAC
So, BCAC = ab
Thus, BC = ak and AC = bk


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ AB2 = AC2- BC2
⇒ AB2= (bk)2 - (ak)2
⇒ AB2 = b2-a2k2
⇒ AB = (b2-a2)k
∴ cos θ  = ABAC = b2-a2kbk = b2-a2b

Page No 352:

Question 34:

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 = (3k)2 + (k)2
⇒ AC2= 4k2
⇒ AC = 2k
∴ sec θ = ACAB = 2kk = 21

Page No 352:

Question 35:

If sec θ = 257, then sin θ = ?
(a) 724
(b) 247
(c) 2425
(d) None of these

Answer:

(c) 2425
Let us first draw a right ABC right angled at B and A=θ.
Given: sec θ = 257
But cos θ1sec θABAC = 725
Thus, AC = 25k and AB = 7k


Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
BC2 = AC2-AB2
BC2 = (25k)2-(7k)2
BC2= 576k2
BC = 24k

∴ sin θ = BCAC = 24k25k = 2425



Page No 353:

Question 36:

If sinθ = 12, then cotθ = ?
(a) 13
(b) 3
(c) 32
(d) 1

Answer:

(b) 3

Given: sinθ = 12, but sinθ = BCAC
So, BCAC = 12
Thus, BC = k and AC = 2k


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
AB2 = AC2- BC2
AB2= (2k)2 - (k)2
AB2 = 3k2
AB = 3k
So, tanθ = BCAB = k3k = 13
∴ cotθ = 1tanθ = 3

Page No 353:

Question 37:

If cosθ = 45, then tanθ = ?
(a) 34
(b) 43
(c) 35
(d) 53

Answer:

(a) 34
Since cosθ = 45 but cosθ = ABAC
So, ABAC = 45
Thus, AB = 4k and AC = 5k


Using Pythagoras theorem in triangle ABC, we have:
AC2 = AB2 + BC2
⇒ BC2 = AC2- AB2
⇒ BC2  = (5k)2 - (4k)2
⇒ BC2 = 9k2
⇒ BC = 3k
∴ tanθ = BCAB = 34

Page No 353:

Question 38:

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]

Page No 353:

Question 39:

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]

Page No 353:

Question 40:

If tan θ = 43, then (sinθ + cosθ) = ?
(a) 73
(b) 74
(c) 75
(d) 57

Answer:

(c) 75
Let us first draw a right ABC right angled at B and A=θ.
tan θ = 43 = BCAB
So, AB = 3k and BC = 4k

Using Pythagoras theorem, we get:
AC2 = AB2 + BC2
AC2 = (3k)2 + (4k)2
AC2 = 25k2
AC= 5k
Thus, sin θ = BCAC = 45
and cos θ =ABAC = 35
∴ (sin θ + cos θ) = ( 45 + 35) = 75

Page No 353:

Question 41:

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

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

Answer:

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)2 = 52
⇒ tan2θ + 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 353:

Question 42:

If (cos θ + sec θ) = 52, then (cos2 θ + sec2 θ) = ?
(a) 214
(b) 174
(c) 294
(d) 334

Answer:

(b) 174
We have (cos θ +sec θ) = 52
Squaring both sides, we get:
(cos θ + sec θ)2 = (52)2
cos2θ + sec2θ + 2 cos θ sec θ = 254
cos2θ + sec2θ + 2 = 254      [∵ sec θ = 1cos θ]
cos2θ + sec2θ = 254 − 2 = 174

Page No 353:

Question 43:

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

Page No 353:

Question 44:

If 7 tan θ = 4, then (7sinθ-3cosθ)(7sinθ+3cosθ) = ?
(a) 17
(b) 57
(c) 37
(d) 514

Answer:

(a) 17

7 tan θ = 4

Now, dividing the numerator and denominator of the given expression by cos θ, we get:
1cosθ7sinθ - 3cosθ1cosθ7sinθ + 3cosθ

= 7tanθ - 37tanθ + 3

= 4 - 34 + 3         [∵ 7 tan θ = 4]
17

Page No 353:

Question 45:

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]



Page No 354:

Question 46:

If θ = ab, then (a sinθ-b cosθ)(a sinθ+b cosθ) = ?
(a) (a2+b2)(a2-b2)
(b) (a2-b2)(a2+b2)
(c) a2(a2+b2)
(d) b2(a2+b2)

Answer:

(b) a2-b2a2 +b2
We have tan θ = ab

Now, dividing the numerator and denominator of the given expression by cos θ, we get:
 asin θ - bcos θasin θ + bcos θ=1cosθasin θ - bcos θ1cosθasin θ + bcos θ=atanθ - b atanθ + b=a2b-ba2b+b=a2 -b2a2 +b2

Page No 354:

Question 47:

If sin A + sin2A = 1, then (cos2A + cos4A) = ?
(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

Page No 354:

Question 48:

If cos A + cos2A = 1, then (sin2A + sin4A) = ?
(a) 1
(b) 2
(c) 4
(d) 3

Answer:

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

Page No 354:

Question 49:

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

Page No 354:

Question 50:

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:

(a) (cosec A − cot A)

1cosA1+cosA=(1cosA)(1+cosA)×(1cosA)(1cosA)            [Multiplying the numerator and denominator by (1cosA)]=(1cosA)(1cosA)1cos2A=1cosA1cos2A=1cosAsin2A=1cosAsinA=1sinAcosAsinA=cosecAcotA

Page No 354:

Question 51:

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)

Page No 354:

Question 52:

(cosec θ − cot θ)2 = ?
(a) 1+cosθ1-cosθ
(b) 1-cosθ1+cosθ
(c)  1+sinθ1-sinθ
(d) None of these

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 354:

Question 53:

(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



Page No 359:

Question 1:

cos256°+cos234°sin256°+sin234°+3tan256° tan234°=?
(a) 312
(b) 4
(c) 6
(d) 5

Answer:

(b) 4

 cos2560+cos2340sin2560+sin2340+3tan2560tan2340={cos(900340)}2+cos2340{sin(900340)}2+sin2340+3{tan(900340)}2tan2340=sin2340+cos2340cos2340+sin2340+3cot2340tan2340      [cos(900θ)=sinθ, sin(900θ)=cosθ and tan(900θ)=cotθ]=11+3×1          [cotθ=1tanθ and sin2θ+cos2θ=1 ]=4

Page No 359:

Question 2:

The value of (sin230°cos245°+4tan230°+12sin290°+18cot260°)=?
(a) 38
(b) 58
(c) 6
(d) 2

Answer:

(d) 2
(sin2300cos2450)+4tan2300+12sin2900+18cot2600=122×1(2)2+4×1(3)2+12×12+18×1(3)2                [sin300=12 and cos450=12 and tan300=12 and cot600=13]=14×12+4×13+12+124=18+43+12+124=3+32+12+124=48242

Page No 359:

Question 3:

If cos A + cos2A = 1, then (sin2A + sin4 A) = ?
(a) 12
(b) 2
(c) 1
(d) 4

Answer:

(c) 1 cosA+2A=1=> cosA=sin2A    ...(i)Squaring both sides of (i), we get:cos2A=sin4A           ...(ii)Adding (i) and (ii), we get:sin2A+sin4A=cosA+cos2A =>sin2A+sin4A=1          [cosA+cos2A=1]



Page No 360:

Question 4:

If sin θ=32, then (cosec θ + cot θ) = ?
(a) (2+3)
(b) 23
(c) 2
(d) 3

Answer:

(d) 3
 Given: sinθ=32 and cosecθ=23cosec2θcot2θ=1=>cot2θ=cosec2θ1 =>cot2θ=431                [Given]=>cotθ=13cosecθ+cotθ=23+13 =33=3×33=3

Page No 360:

Question 5:

If cot A=45, prove that (sinA+cosA)(sinA-cosA)=9.

Answer:

Given: cotA=45Writing cot A = cos Asin A and sqauring the equation, we get:  cos2Asin2A=1625=>25cos2A=16sin2A=>25cos2A=1616cos2A=>cos2A=1641=>cosA=441sin2A=1cos2A=11641Now, sinA=2541=>sinA=541LHS=sinA+cosAsinAcosA =541+441541441=91=9=RHS

Page No 360:

Question 6:

If 2x = sec A and 2x = tan A, prove that x2-1x2=14.

Answer:

 Given: 2x=secA=>x=secA2      ...(i)and 2x=tanA=>1x=tanA2     ...(ii)x+1x=secA2+tanA2             [From (i) and (ii)] Also, x1x=secA2tanA2(x+1x)(x1x)=(secA2+tanA2)(secA2tanA2)=>x21x2=14(sec2Atan2A)x21x2=14×1                 (sec2Atan2A=1)=14 Hence proved.

Page No 360:

Question 7:

If 3 tan θ = 3 sin θ, prove that (sin2 θ − cos2 θ) = 13.

Answer:

 Given: 3tanθ=3sinθ=>3cosθ=3           [tanθ=sinθcosθ]=>cosθ=33=>cos2θ=39sin2θ=1 39=> sin2θ=69LHS=sin2θcos2θ=6939         [sin2θ=69,cos2θ=39]=39=13=RHSHence proved.

Page No 360:

Question 8:

Prove that (sin273°+sin217°)(cos228°+cos262°)=1.

Answer:

(sin273°+sin217°)(cos228°+cos262°)=1.

 LHS=sin2730+sin2170cos2280+cos2620=[sin(900170)]2+sin2170[cos(900620)]2+cos2620=cos2170+sin2170sin2620+cos2620=11                       [sin2θ+cos2θ=1]=1=RHS

Page No 360:

Question 9:

If 2 sin 2θ =3, prove that θ = 30°.

Answer:

  2sin(2θ)=3=>sin(2θ)=32=>sin(2θ)=sin(600)=>2θ=600=>θ=6002=>θ=300

Page No 360:

Question 10:

Prove that 1+cos A1-cos A = (cosec A + cot A).

Answer:

1+cos A1-cos A= (cosec A + cot A).

LHS = 1+cosA1cosAMultiplying the numerator and denominator by (1+cosA), we have:(1+cosA)2(1cosA)(1+cosA)=(1+cosA)21cos2A=1+cosAsin2A=1+cosAsinA=1sinA+cosAsinA=cosecA+cotA=RHS Hence proved.

Page No 360:

Question 11:

If cosec θ + cot θ = p, prove that cos θ = (p2-1)(p2+1).

Answer:

  cosecθ+cotθ=p=>1sinθ+cosθsinθ=p=>1+cosθsinθ=pSquaring both sides, we get:(1+cosθsinθ)2=p2=>(1+cosθ)2sin2θ=p2=>(1+cosθ)21cos2θ=p2=>(1+cosθ)2(1+cosθ)(1cosθ)=p2=>(1+cosθ)(1cosθ)=p2=>1+cosθ=p2(1cosθ)=>1+cosθ=p2p2cosθ=>cosθ(1+p2)=p21=>cosθ=p21p2+1Hence proved.

Page No 360:

Question 12:

Prove that (cosec A − cot A)2 = (1-cos A)(1+cos A).

Answer:

(cosec A − cot A)2 = (1-cos A)(1+cos A).

 LHS=(cosecAcotA)2=(1sinAcosAsinA)2=(1cosAsinA)2=(1cosA)2sin2A=(1cosA)21cos2A             [sin2θ+cos2θ=1]=(1cosA)(1cosA)(1cosA)(1+cosA)=(1cosA)(1+cosA)= RHSHence proved.

Page No 360:

Question 13:

If 5 cot θ = 3, find the value of 5sinθ-3cosθ4sinθ+3cosθ.

Answer:

 Given: 5cotθ=3=>5cosθsinθ=3         [cotθ=cosθsinθ]=>5cosθ=3sinθSquaring both sides, we get:25cos2θ=9sin2θ=>25cos2θ=99cos2θ            [sin2θ+cos2θ=1]=>34cos2θ=9=>cosθ=934=>cosθ=334Again, sin2θ=1cos2θ=>sin2θ=34934=2534=>sinθ=534LHS=(5sinθ3cosθ4sinθ+3cosθ)=5×5343×3344×534+3×334              [cosθ=334,sinθ=534]=25920+9=1629

Page No 360:

Question 14:

Prove that (sin 32° cos 58° + cos 32° sin 58°) = 1.

Answer:

(sin 32° cos 58° + cos 32° sin 58°) = 1

LHS=sin320cos580+cos320sin580=sin(900580)cos580+cos(900580)sin580=cos580×cos580+sin580×sin580           [sin(900θ)=cosθ,cos(900θ)=cosθ]   =cos2580+sin2580=1                     [sin2θ+cos2θ=1]=RHS

Page No 360:

Question 15:

If x = a sin θ + b cos 0 and y = a cos 0 b sin θ, prove that x2+y2=a2+b2.

Answer:

Given: x=asinθ+bcosθSquaring both sides, we get: x2=a2sin2θ+2absinθcosθ+b2cos2θ     ...(i)Also, y=acosθbsinθSquaring both sides, we get:y2=a2cos2θ2absinθcosθ+b2sin2θ    ...(ii)LHS= x2+ y2=a2sin2θ+2absinθcosθ+b2cos2θ+    a2cos2θ2absinθcosθ+b2sin2θ     [using (i)and (ii)]=a2(sin2θ+cos2θ)+b2(sin2θ+cos2θ)=a2+b2                  [sin2θ+cos2θ=1]=RHSHence proved.

Page No 360:

Question 16:

Prove that (1+sinθ)(1-sinθ) = (sec θ + tan θ)2.

Answer:

(1+sinθ)(1-sinθ)= (sec θ + tan θ)2

LHS=(1+sinθ)(1sinθ)Multiplying the numerator and denominator by (1+sinθ), we get:(1+sinθ)21sin2θ=1+2sinθ+sin2θcos2θ                   [sin2θ+cos2θ=1]=sec2θ+2×sinθcosθ×secθ+tan2θ=sec2θ+2×tanθ×secθ+tan2θ=(secθ+tanθ)2=RHSHence proved.

Page No 360:

Question 17:

Prove that 1(secθ-tanθ)-1cosθ=1cosθ-1(secθ+tanθ).

Answer:

1(secθ-tanθ)-1cosθ=1cosθ-1(secθ+tanθ)

LHS=   1secθtanθ1cosθ=(secθ+tanθ)(secθtanθ)(secθ+tanθ)secθ            Multipying the numerator and denominator by (secθ+tanθ) =secθ+tanθsec2θtan2θsecθ=secθ+tanθsecθ                   [sec2θtan2θ=1]=tanθRHS=1cosθ1secθ+tanθ=secθ(secθtanθ)sec2θtan2θ                             Multipying the numerator and denomenator by (secθtanθ)  =secθ+tanθsecθ              [sec2θtan2θ=1]=tanθLHS=RHSHence Proved

Page No 360:

Question 18:

Prove that (sin A-2sin3A)(2cos3A-cos A)=tan A.

Answer:

LHS=(sinA2sin3A)(2cos3AcosA)=sinA(12sin2A)cosA(2cos2A1)=tanA{(sin2A+cos2A2sin2A)2cos2Asin2Acos2A}              [sin2A+cos2A=1]=tanA{(cos2Asin2A)(cos2Asin2A)}=tanA=RHS

Page No 360:

Question 19:

Prove that tan A(1-cot A)+cot A(1-tan A)=(1+tan A+cot A).

Answer:

LHS=tanA(1cotA)+cotA(1tanA)=tanA(1cotA)+cot2A(cotA1)            [tanA=1cotA]=tanA(1cotA)cot2A(1cotA)=tanAcot2A(1cotA)=(1cotA)cot2A(1cotA)=1cot3AcotA(1cotA)=(1cotA)(1+cotA+cot2A)cotA(1cotA)               [a3b3=(ab)(a2+ab+b2)] =1cotA+cot2AcotA+cotAcotA=1+tanA+cotA=RHSHence proved

Page No 360:

Question 20:

If sec 5 A = cosec (A − 36) and 5 A is an acute angle, find the value of A.

Answer:

Given: sec5A=cosec(A360)=> cosec(9005A)=cosec(A360)         [cosec(900θ)=secθ]=> 9005A=A360=>6A=900+360 => 6A=1260=>  A=210 



Page No 602:

Question 11:

If A and B are acute angles such that sin A = cos B then (A + B) = ?
(a) 45°
(b) 60°
(c) 90°
(d) 180°

Answer:

Given that and are acute angles such that sin A = cos B.
sinA=sin90°-BA=90°-B A+B=90°.
Hence, the correct answer is option C.



View NCERT Solutions for all chapters of Class 10