Rs Aggarwal 2019 Solutions for Class 10 Math 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 Math Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 583:

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 583:

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 583:

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 583:

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 583:

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 583:

Answer:

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



Page No 584:

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 584:

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 584:

Answer:

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

Page No 584:

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 584:

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 584:

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 584:

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 584:

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 584:

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 584:

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 584:

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 584:

Answer:

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

Page No 584:

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 584:

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 585:

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 585:

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 585:

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 585:

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 585:

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 585:

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 585:

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 585:

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 585:

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 585:

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 585:

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 585:

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 585:

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 585:

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 586:

Answer:

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

Hence, LHS = RHS

Page No 586:

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 586:

Answer:

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

Page No 586:

Answer:

 1+sin2θ=3sinθcosθDivide by cos2θ throughout to get: sec2θ+tan2θ=3tanθ1+tan2θ+tan2θ=3tanθ2tan2θ-3tanθ+1=0, which is quadratic in tanθ.Solving using Quadratic formula, we get:tanθ=+3±-32-42122=+3±9-84=+3±14tanθ=+44,+24 =1 or 12.



Page No 594:

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 594:

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 594:

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 594:

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 594:

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 594:

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 594:

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 594:

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 594:

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 594:

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 594:

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 594:

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 594:

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 595:

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 595:

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 596:

Answer:

1-sin2θsec2θ=cos2θ×1cos2θ=1

Page No 596:

Answer:

1-cos2θcosec2θ=sin2θ×1sin2θ=1

Page No 596:

Answer:

1+tan2θcos2θ=sec2θ×1sec2θ=1

Page No 596:

Answer:

1+cot2θsin2θ=cosec2θ×1cosec2θ=1

Page No 596:

Answer:

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

Page No 596:

Answer:

cot2θ-1sin2θ=cot2θ-cosec2θ=-1

Page No 596:

Answer:

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

Page No 596:

Answer:

cosec290°-θ-tan2θ=sec2θ-tan2θ=1



Page No 597:

Answer:

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

Page No 597:

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 597:

Answer:

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

Page No 597:

Answer:

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

Page No 597:

Answer:

3cot2θ-3cosec2θ=3cot2θ-cosec2θ=3-1=-3

Page No 597:

Answer:

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

Page No 597:

Answer:

tan2θ-sec2θcot2θ-cosec2θ=-1-1=1

Page No 597:

Answer:

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

Now,

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

Page No 597:

Answer:

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

Page No 597:

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 597:

Answer:

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

Page No 597:

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 597:

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 597:

Answer:

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

Page No 597:

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 597:

Answer:

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

Page No 597:

Answer:

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

Page No 597:

Answer:

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

Page No 597:

Answer:

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

Page No 597:

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 597:

Answer:

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

Page No 597:

Answer:

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

Page No 597:

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 597:

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 597:

Answer:

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



Page No 598:

Answer:

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

Page No 598:

Answer:

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

Page No 598:

Answer:

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

Page No 598:

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 598:

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 598:

Answer:

cotθ=cosθsinθ=1-sin2θsinθ=1-x22

Page No 598:

Answer:

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



Page No 601:

Answer:

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

Page No 601:

Answer:

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

Page No 601:

Answer:

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

Page No 601:

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 601:

Answer:

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

Page No 601:

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 601:

Answer:

(c) 1We have:    (sin470cos430+cos470sin430)=sin470cos(900470)+cos470sin(900470)=sin470sin470+cos470cos470              [cos(900θ)=sinθ and sin(900θ)=cosθ]=sin2470+cos2470=1



Page No 602:

Answer:

(d) 2We have:      sec700sin200+cos200cosec700=sin200cos700+cos200sin700=sin200cos(900200)+cos200sin(900200)=sin200sin200+cos200cos200                        [cos(900θ)=sinθ and sin(900θ)=cosθ]=1+1=2

Page No 602:

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 602:

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 602:

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.

Page No 602:

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 602:

Answer:

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

Page No 602:

Answer:

sec210°-cot280°=sec210°-cot290°-10°=sec210°-tan210°=1.
Hence, the correct answer is option A.

Page No 602:

Answer:

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

Page No 602:

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 602:

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 602:

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 602:

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 602:

Answer:

(a) 30o

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

Page No 602:

Answer:

(c) 20o

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



Page No 603:

Answer:

(b) 30o

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

Page No 603:

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 603:

Answer:

(a) 1

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

Page No 603:

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 603:

Answer:

(b) 3

sec260o − 1 = (2)2 − 1 = 4 − 1 = 3

Page No 603:

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 603:

Answer:

(b) 14

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

Page No 603:

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 603:

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 603:

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 = 10 k


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

Page No 603:

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 603:

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 603:

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



Page No 604:

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 604:

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 604:

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 604:

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 604:

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 604:

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 604:

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 604:

Answer:

(b) 174
We have (cos θ +sec θ) = 52
Squaring both sides, we get:
(cos θ + sec θ)2 = (52)2
cos2 θ + sec2 θ + 2 cos θ sec θ = 254
cos2 θ + sec2 θ + 2 = 254      [∵ sec θ = 1cos θ]
cos2 θ + sec2 θ = 254 − 2 = 174

Page No 604:

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 604:

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 605:

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 605:

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 605:

Answer:

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

Page No 605:

Answer:

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

Page No 605:

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 605:

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 605:

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 605:

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 605:

Answer:

(b) cos A

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



Page No 610:

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 610:

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 610:

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 611:

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 611:

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 611:

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 611:

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 611:

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 611:

Answer:

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

Page No 611:

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 611:

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 611:

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 611:

Answer:

Given: 5cotθ=3 cotθ=35=Base(B)Perpendicular(P)Using Pythagoras Theorem, Hypotenuse(H)=B2+P2=32+52=34sinθ=PH=534, cosθ=BH=334Now, 5sinθ-3cosθ4sinθ+3cosθ=5×534-3×3344×534+3×334=2534-9342034+934=16342934=1634×3429=1629

Page No 611:

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 611:

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 611:

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 611:

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 611:

Answer:

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

Page No 611:

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 611:

Answer:

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



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