Rs Aggarwal 2015 Solutions for Class 10 Math Chapter 7 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 2015 Book of Class 10 Math Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2015 Solutions. All Rs Aggarwal 2015 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 216:

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 216:

Question 34:

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 216:

Question 35:

(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 314:

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 314:

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 314:

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 314:

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 314:

Question 5:

(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 314:

Question 6:

(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 314:

Question 7:

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 314:

Question 8:

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 314:

Question 9:

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 314:

Question 10:

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 314:

Question 11:

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 314:

Question 12:

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 314:

Question 13:

(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 315:

Question 14:

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 315:

Question 15:

(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 315:

Question 16:

(i) 1-sinθ1+sinθ=(secθ-tanθ)2
(ii) 1+cosθ1-cosθ=(cosecθ+cotθ)2

Answer:

(i) LHS=1sinθ1+sinθ              =(1sinθ)(1sinθ)(1+sinθ)(1sinθ)     [Dividing the numerator and denominator by (1sinθ)]              =(1sinθ)2(1sin2θ)              =1+sin2θ2sinθcos2θ              =1cos2θ+sin2θcos2θ2sinθcos2θ              =sec2θ+tan2θ2×1cosθ×sinθcosθ              =sec2θ+tan2θ2secθtanθ              =(secθ-tanθ)2              =RHS Hence, LHS= RHS(ii) LHS=1+cosθ1cosθ              =(1+cosθ)(1+cosθ)(1cosθ)(1+cosθ)         [Dividing the numerator and denominator by (1+cosθ)]              =(1+cosθ)2(1cos2θ)              =1+cos2θ+2cosθsin2θ              =1sin2θ+cos2θsin2θ+2cosθsin2θ              =cosec2θ+cot2θ+21sinθ×cosθsinθ              =cosec2θ+cot2θ+2cosec2θcot2θ              =(cosecθ+cotθ)2              =RHS Hence, LHS = RHS

Page No 315:

Question 17:

1+tan2θ1+cot2θ=1+tanθ1-cotθ2=tan2θ

Answer:

Here, LHS=1+tan2θ1+cot2θ         =sec2θcosec2θ        =1cos2θ1sin2θ        =sin2θcos2θ        =tan2θAgain, LHS = 1tanθ1cotθ2        =1sinθcosθ1cosθsinθ2       ={(cosθsinθ)cosθ×sinθ(sinθcosθ)}2       ={(sinθcosθ)cosθ×sinθ(sinθcosθ)}2       =sinθcosθ2       =sin2θcos2θ       =tan2θ

∴ LHS = RHS
Hence proved.

Page No 315:

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 315:

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 315:

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 315:

Question 21:

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 315:

Question 22:

(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 315:

Question 23:

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 315:

Question 24:

(1-sinθ+cosθ)2=2(1+cosθ)(1-sinθ)

Answer:

LHS=(1sinθ+cosθ)2         ={(1sinθ)+cosθ}2         =(1sinθ)2+cos2θ+2(1sinθ)cosθ         =1+sin2θ2sinθ+cos2θ+2cosθ2sinθcosθ         =1+12sinθ+2cosθ2sinθcosθ         =22sinθ+2cosθ2sinθcosθ         =2{1(1sinθ)+cosθ(1sinθ)}         =2(1+cosθ)(1sinθ)         =RHS

Hence, LHS = RHS

Page No 315:

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 316:

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 316:

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 316:

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 316:

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 316:

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 316:

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 316:

Question 32:

cot2θ(secθ-1)(1+sinθ)=sec2θ.1-sinθ1+secθ

Answer:

LHS=cot2θ(secθ1)(1+sinθ)         =cos2θsin2θ(1cosθ1)(1+sinθ)          =cosθsin2θcos2θsin2θ(1+sinθ)         =cosθ(1cosθ)sin2θ(1+sinθ)         =cosθ(1cosθ)(1cos2θ)(1+sinθ)         =cosθ(1cosθ)(1+cosθ)(1cosθ)(1+sinθ)         =cosθ(1+1secθ)(1+sinθ)        =cosθsecθ(secθ+1)(1+sinθ)        =1(secθ+1)(1+sinθ)        =(1sinθ)(secθ+1)(1+sinθ)(1sinθ)            [Multiplying the numerator and denominator by (1-sinθ)]        =(1sinθ)(secθ+1)(1sin2θ)       =(1sinθ)(secθ+1)cos2θ       =sec2θ(1sinθ)(secθ+1)       =RHS
Hence, LHS = RHS

Page No 316:

Question 36:

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 316:

Question 37:

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 317:

Question 38:

Show that each of the following is an identity:
(i) tan2θ(1+tan2θ)=sin2θ
(ii) cotθ+cosθcotθ-cosθ=1+sinθ1-sinθ

Answer:

(i) LHS=tan2θ(1+tan2θ)             =tan2θsec2θ             =sin2θcos2θ1cos2θ             =sin2θ             =RHSi.e., LHS=RHSThis is an identity.(ii) LHS=cotθ+cosθcotθcosθ              =cosθsinθ+cosθcosθsinθcosθ              =cosθ+cosθsinθcosθcosθsinθ              =cosθ(1+sinθ)cosθ(1sinθ)              =1+sinθ1sinθ              =RHSi.e., LHS=RHSThis is an identity.



Page No 321:

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 321:

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 322:

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 322:

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 322:

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 322:

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 322:

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 322:

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 322:

Question 9:

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+tanθ)2(1cosθ-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 322:

Question 10:

If (cos θ + sin θ) = 1, prove that (cos θ − sin θ) = ± 1.

Answer:

We have (cosθ+sinθ)=1We know that (cosθ+sinθ)2+(cosθsinθ)2=2(sin2θ+cos2θ)=>(1)2+(cosθsinθ)2=2×1=>1+(cosθsinθ)2=2=>(cosθsinθ)2=1=>cosθsinθ=±1∴ cosθsinθ=±1

Page No 322:

Question 11:

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 322:

Question 12:

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 322:

Question 13:

If a cos3θ + 3a sin2θ cos θ = m and a sin3θ + 3a sin θ cos2θ = n, prove that
(m + n)2/3 + (mn)2/3 = 2a2/3.

Answer:

We have acos3θ+3asin2θcosθ=mand asin3θ+3asinθcos2θ=nNow, (m+n)=acos3θ+3asin2θcosθ+asin3θ+3asinθcos2θ=a[(cos3θ+sin3θ)+3sinθcosθ(sinθ+cosθ)](m+na)=[(cos3θ+sin3θ)+3sinθcosθ(sinθ+cosθ)]m+n=a(cosθ+sinθ)3(m+n)23=(cosθ+sinθ)2a23                  ...(i)Similarly, (mn)23=(cosθsinθ)2a23                          ...(ii)Adding equation (i) and (ii), we get: (m+n)23+(mn)23=a23[cos2θ+sin2θ+2sinθcosθ+cos2θ+sin2θ2sinθcosθ]=a23[2]=2a23

Page No 322:

Question 14:

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 322:

Question 15:

If sin θ + cos θ = 2 sin (90° − θ), show that cot θ = 2+1.

Answer:

Given: cosθ+sinθ=2sin(900θ)=>cosθ+sinθ=2cosθ      {∵ sin(90°-θ) = cosθ}=>cosθ+sinθsinθ=2cosθsinθ    (Dividing both sides by sinθ)=>cotθ+1=2cotθ =>(21)cotθ=1=>cotθ=1(21) =>cotθ=1(21)×(2+1)(2+1)=>cotθ=(2+1)21        =>cotθ=(2+1)Hence proved.

Page No 322:

Question 16:

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 322:

Question 17:

If tan θ = ab, show that a sinθ-b cosθa sinθ+b cosθ=a2-b2a2+b2.

Answer:

Given: tanθ=abLHS=asinθbcosθasinθ+bcosθ =atanθbatanθ+b             (Dividing both numerator anddenominator by cosθ) =a×abba×ab+b =a2bba2b+b =a2b2a2+b2

Hence, LHS = RHS



Page No 323:

Question 1:

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 324:

Question 2:

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 324:

Question 3:

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

Answer:

(b) cos A

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

Page No 324:

Question 4:

(1 + tan A + sec A)(1 + cot − cosec A) = ?
(a) 1
(b) 0
(c) 2
(d) 4

Answer:

(c) 2

(1+tanA+secA)(1+cotAcosecA)=(1+sinAcosA+1cosA)(1+cosAsinA1sinA)=(cosA+sinA+1cosA)(sinA+cosA1sinA)={(cosA+sinA)+1}{(sinA+cosA)1}cosAsinA=(cosA+sinA)21cosAsinA=(cos2A+sin2A+2cosAsinA)1cosAsinA=1+2cosAsinA1cosAsinA  (cos2A+sin2A=1)=2cosAsinAcosAsinA=2

Page No 324:

Question 5:

(1+tan2A)(1+cot2A)=?
(a) tan2 A
(b) cot2 A
(c) sec2 A
(d) cosec2 A

Answer:

(a) tan2A
1+tan2A1+cot2A=1+sin2Acos2A1+cos2Asin2A=cos2A+sin2Acos2Asin2A+cos2Asin2A=1cos2A1sin2A                             (cos2A+sin2A=1)=sin2Acos2A=tan2A

Page No 324:

Question 6:

(sin A + cos A)2 + (sin A − cos A)2 = ?
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(c) 2
(sinA+cosA)2+(sinAcosA)2=(sin2A+2sinAcosA+cos2A)+(sin2A+cos2A2sinAcosA)=2sin2A+2cos2A=2(sin2A+cos2A) =2×1        (sin2A+cos2A=1)=2

Page No 324:

Question 7:

(sec4A − sec2A) = ?
(a) tan4A − tan2A
(b) tan2A + tan4A
(c) tan2A − tan4A
(d) None of these

Answer:

(b) tan2A + tan4A

(sec4Asec2A)=sec2A(sec2A1)=sec2A(tan2A)=(1+tan2A)(tan2A)=tan2A+tan4A

Page No 324:

Question 8:

(cos4θ − sin4θ)=?
(a) 1 − 2 sin2θ
(b) 1 − 2 cos2θ
(c) 2 − sin2θ
(d) 2 − cos2θ

Answer:

a) 1 − 2 sin2θ

(cos4θsin4θ)=(cos2θ)2(sin2θ)2=(cos2θ+sin2θ)(cos2θsin2θ)=1(1sin2θsin2θ)=12sin2θ

Page No 324:

Question 9:

sinθ(1+cosθ)=?
(a) cosθ(1-sinθ)
(b) (1-cosθ)sinθ
(c) (1-sinθ)cosθ
(d) None of these

Answer:

(b) (1-cosθ)sinθ

sinθ1+cosθ=sinθ(1cosθ)(1+cosθ)(1cosθ)        [Multiplying the numerator and denominator by (1cosθ)]=sinθ(1cosθ)1cos2θ=sinθ(1cosθ)sin2θ=(1cosθ)sinθ

Page No 324:

Question 10:

sinθ(1+cosθ)+sinθ(1-cosθ)=?
(a) 2 sin θ
(b) 2 cos θ
(c) 2 sec θ
(d) 2 cosec θ

Answer:

(d) 2 cosec θ

(sinθ1+cosθ)+(sinθ1cosθ)=sinθ(1cosθ)+sinθ(1+cosθ)(1+cosθ)(1cosθ)=sinθsinθcosθ+sinθ+sinθcosθ1cos2θ=2sinθsin2θ              [1cos2θ=sin2θ]=2sinθ=2cosecθ

Page No 324:

Question 11:

sinθ(1-cotθ)+cosθ(1-tanθ)=?
(a) (cos θ + sin θ)
(b) (cos θ − sin θ)
(c) 0
(d) 2 tan θ

Answer:

(a) (cos θ + sin θ)

sinθ(1cotθ)+cosθ(1tanθ)=sinθ1cosθsinθ+cosθ1sinθcosθ=(sinθsinθcosθsinθ)+(cosθcosθsinθcosθ)=(sin2θsinθcosθ)+(cos2θcosθsinθ)=(cos2θcosθsinθ)(sin2θcosθsinθ)=(cos2θsin2θcosθsinθ)={(cosθ+sinθ)(cosθsinθ)(cosθsinθ)}=cosθ+sinθ

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Question 12:

(1-tan2θ)(1+tan2θ)=?
(a) (sin2θ − cos2θ)
(b) (cos2θ − sin2θ)
(c) (cot2θ − tan2θ)
(d) (tan2θ − cot2θ)

Answer:

(b) (cos2θ − sin2θ)

1tan2θ1+tan2θ=(1Sin2θCos2θ)(1+Sin2θCos2ϑ)=(Cos2θSin2θCos2θ)(Cos2θ+Sin2θCos2θ)=cos2θsin2θcos2θ+sin2θ=cos2θsin2θ            [cos2θ+sin2θ=1]



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Question 13:

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

Answer:

(a) (sec A + tan A)

1+sinA1sinA=(1+sinA)(1sinA)×(1+sinA)(1+sinA)         [Multiplying and dividing by (1+sinA)]=(1+sinA)1sin2A=(1+sinA)cos2A=(1+sinA)cosA=1cosA+sinAcosA=secA+tanA

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Question 14:

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

Answer:

(b) (sec A − tan A)

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

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Question 15:

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

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Question 16:

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:

(b) (cosec A + cot A)

1+cosA1cosA=(1+cosA)(1cosA)×(1+cosA)(1+cosA)           [Multiplying the numerator and denominator by (1+cosA)]=(1+cosA)(1+cosA)1cos2A=1+cosA1cos2A=1+cosAsin2A=1+cosAsinA=1sinA+cosAsinA=cosecA+cotA

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Question 17:

sec A-tan A sec A+tan A=?
(a) (sec A − tan A)
(b) (sec A + tan A)
(c) sec A tan A
(d) None of these

Answer:

(a) (sec A − tan A)

secAtanAsecA+tanABy rationalising, we get:(secAtanA)(secA+tanA)×(secAtanA)(secAtanA) =(secAtanA)(secAtanA)(sec2Atan2A)=(secAtanA)1   [sec2Atan2A=1] =secAtanA

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Question 18:

(sin4θ − cos4θ + 1) cosec2θ = ?
(a) 4
(b) 3
(c) 2
(d) 1

Answer:

(c) 2

(sin4θcos4θ+1)cosec2θ={(sin2θcos2θ)(sin2θ+cos2θ)+1}1sin2θ=[{(sin2θcos2θ)×1}+1]1sin2θ=sin2θ+(1cos2θ)sin2θ=sin2θ+sin2θsin2θ                   [(1cos2θ=sin2θ)]=2sin2θsinθ=2

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Question 19:

If x = a cos θ and y = b sin θ, then (b2x2 + a2y2) = ?
(a) a2 + b2
(b) a2b2
(c) ab
(d) a4b4

Answer:

(b) a2b2

x=acosθ and y=bsinθ=>cosθ=xa and sinθ=ybWe know that cos2θ+sin2θ=1=>(xa)2+(yb)2=1=>x2a2+y2b2=1=>x2b2+y2a2a2b2=1=>b2x2+a2y2=a2b2

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Question 20:

If x = a sec θ cos Ï•, y = b θ sin Ï• and z = c tan θ, then x2a2+y2b2=?
(a) 1+z2c2
(b) 1-z2c2
(c) z2c2-1
(d) z2c2

Answer:

(a) 1+z2c2

x=asecθcosϕ, y=bsecθsinϕ and z=ctanθsecθcosϕ=xa, secθsinϕ=yb and tanθ=zcNow, sec2θcos2ϕ+sec2θsin2ϕ=(xa)2+(yb)2    [Squaring both sides]=>sec2θ(cos2ϕ+sin2ϕ)=x2a2+y2b2=>sec2θ=x2a2+y2b2=>1+tan2θ=x2a2+y2b2=>1+z2c2=x2a2+y2b2(x2a2+y2b2)=(1+z2c2)

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Question 21:

secθ-1secθ+1+secθ+1secθ-1=?
(a) 2 sin θ
(b) 2 cos θ
(c) 2 cosec θ
(d) 2 sec θ

Answer:

(c) 2 cosec θ

secθ1secθ+1+secθ+1secθ1 By rationalising, we get:(secθ1)(secθ1)(secθ+1)(secθ1)+(secθ+1)(secθ+1)(secθ1)(secθ+1)=(secθ1)(secθ1)(sec2θ1)+(secθ+1)(secθ+1)(sec2θ1)=(secθ1)2tan2θ+(secθ+1)2tan2θ=(secθ1)tanθ+(secθ+1)tanθ=secθ1+secθ+1tanθ=2secθtanθ=21cosθsinθcosθ=2cosecθ



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Question 22:

If (sin θ + cos θ) = p and (sec θ + cosec θ) = q, then q(p2 − 1) = ?
(a) 2
(b) 2p
(c) 1p2
(d) 1p

Answer:

(b) 2p

Given: sinθ+cosθ = pSquaring both sides, we get: (sinθ+cosθ)2=p2         ...(i)and (secθ+cosecθ)=q        ...(ii)From (i), (sin2θ+cos2θ)+2sinθcosθ=p2 =>1+2sinθcosθ=p2=>2sinθcosθ=p21=>sinθcosθ=p212        ...(iii)According to the question, we have:(secθ+cosecθ)=q=>(1cosθ+1sinθ)=q=>(sinθ+cosθcosθsinθ)=qHence, (pp212)=q=>2pp21=q =>q(p21)=2p

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Question 23:

If (cos θ + sin θ) = 2 cos θ, then (cos θ − sin θ) = ?
(a) 2 sin θ
(b) 2 sec θ
(c) 2 cosec θ
(d) None of these

Answer:

(a) 2 sin θ

Given: (cosθ+sinθ)=2cosθWe have:(cosθ+sinθ)2+(cosθsinθ)2=2(cos2θ+sin2θ)=>(2cosθ)2+(cosθsinθ)2=2×1=>(cosθsinθ)2=22cos2θ=>(cosθsinθ)2=2(1cos2θ)=>(cosθsinθ)2=2sin2θ(cosθsinθ)=2sinθ

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Question 24:

(sinθ-2sin3θ)(2cos3θ-cosθ)=?
(a) tan θ
(b) cot θ
(c) sec θ
(d) cosec θ

Answer:

(a) tan θ
(sinθ2sin3θ2cos3θcosθ)=sinθ(12sin2θ)cosθ(2cos2θ1)=sinθcosθ×cos2θcos2θ           [(12sin2θ)=cos2θ    (2cos2θ1)=cos2θ ] =tanθ

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Question 25:

(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)

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Question 26:

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

Answer:

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

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Question 27:

If (sin θ + cos θ) = 3, then (tan θ + cot θ) = ?
(a) 13
(b) 12
(c) 1
(d) 32

Answer:

(c) 1

(sinθ+cosθ)=3=>(sinθ+cosθ)2=3   [Squaring both sides]=>sin2θ+cos2θ+2sinθcosθ=3=>1+2sinθcosθ=3=>2sinθcosθ=2=>sinθcosθ=1sinθcosθ=sin2θ+cos2θ=>sin2θ+cos2θsinθcosθ=1=>sinθcosθ+cosθsinθ=1=>tanθ+cotθ=1

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Question 28:

If (sin θ + cos θ) = a and (sin3θ + cos3θ) = b, then (3a − 2b) = ?
(a) a3
(b) b3
(c) 0
(d) 1

Answer:

(a) a3

Given: (sinθ+cosθ)=a(sinθ+cosθ)2=a2    [Squaring both sides](sin2θ+cos2θ)+2sinθcosθ=a21+2sinθcosθ=a2sinθcosθ=a212Again, (sinθ+cosθ)3=a3         [Cubing both sides]sin3θ+cos3θ+3sinθcosθ(sinθ+cosθ)=a3b+3a212a=a3 b+3a(a21)2=a32b+3a(a21)=2a32b+3a33a=2a33a+2b=a33a2b=a3



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