ML Aggarwal Solution Class 10 Chapter 18 Trigonometric Identities Exercise 18

 Exercise 18

Question 1

If A is an acute angle and sin A =35  find all other trigonometric ratios of angle A (using trigonometric identities).

Sol :
sinA=35

In ∆ABC, ∠B = 90°

AC = 5 and BC = 3

Figure to be added

AB=AC2BC2=5232

=259=16=4

Now, cosθ=ABAC=45

tanθ=BCAB=34

cotθ=1tanθ=43

secθ=1cosθ=54

cosecθ=1sinθ=53


Question 2

If A is an acute angle and sec A=178 , find all other trigonometric ratios of angle A (using trigonometric identities).

Sol :
secA=178(A is an acute angle )

In right ∆ABC

secA=ACAB=178

AC = 17, AB = 8

BC=AC2AB2=17282
=28964=225=15

Figure to be added

Now sinA=BCAC=1517

cosA=1secA=817

tanA=BCAB=158

cotA=1tanA=815

cosecA=1sinA=1715


Question 3

Express the ratios cos A, tan A and sec A in terms of sin A.

Sol :

cosA=1sin2A

tanA=sinAcosA=sinA1sin2A

secA=1cosA=11sin2A


Question 4

If tanA=13, find all other trigonometric ratios of angle A.

Sol :
tanA=13

In right ∆ABC,

tanA=BCAB=13

BC=1,AB=3

AC=AB2+BC2=(3)2+(1)2

=3+1=4=2

Figure to be added

sinA=BCAC=12

cosA=ABAC=32

cotA=1tanA=3

secA=1cosA=23

cosecA=1sinA=21=2


Question 5

If 12 cosec θ = 13, find the value of 2sinθ3cosθ4sinθ9cosθ

Sol :

12 cosec θ = 13

cosecθ=1312

In right ∆ABC,

∠A = θ

cosecθ=ACBC=1312

Figure to be added

AC=13,BC=12

AB=AC2BC2=132122

=169144=25=5


Now sinθ=BCAC=1213

cosθ=ABAC=513

Now 2sinθ3cosθ4sinθ9cosθ=2×12133×5134×12139×513

=2413151348134513

=913313=913×133=3


Without using trigonometric tables, evaluate the following (6 to 10) :


Question 6

(i) cos226+cos64sin26+tan36cot54

(ii) sec17cosec73+tan68cot22+cos244+cos246

Sol :

Given that

(i) cos226+cos64sin26+tan36cot54

cos226+cos(9026)sin26+tan36cot(9036)

=cos226+sin226+tan36tan36

=1+1=2

[cos(90θ)=sinθ and cot(90θ)=tanθ,sin2θ+cos2θ=1]


(ii) sec17cosec73+tan68cot22+cos244+cos246

sec17cosec73+tan68cot22+cos244+cos246

=sec(9073)cosec(73)+tan(9022)cot22+cos2(9046+cos246

=cosec73cosec73+cot22cot22+sin246+cos246 (sin2θ+cos2θ=1)

=1+1+1=3


Question 7

(i) sin65cos25+cos32sin58sin28sec62+cosec230(2015)

(ii) sin29cosec61+2cot8cot17cot45cot73cot823(sin238+sin252)

Sol :

Given :

(i) sin65cos25+cos32sin58sin28sec62+cosec230

=sin65cos(9065)+cos32sin(9032)sin28×sec(9028)+(2)2

=sin65sin65+cos32cos32sin28×cosec28+4 {sin(90+θ)=cosθcos(90θ)=sinθ}

=1+1-1+4

=6-1=5


(ii) sin29cosec61+2cot8cot17cot45cot73cot823(sin238+sin252)

=sec29cosec(9029)+2cot8cot82cot17cot73cot453(sin238+sin2(9038)

sec29sec29+2cot8cot(908)cot17cot(9017)cot453(sin238+cos238)

=1+2cot8tan8cot17cot17cot453(1)

{sec(90θ)=cosecθ,cotθ=tan(90θ) and sin2θ+cos2θ=1}

=1+2×1×1×13(cot45=1)

=1+2-3=0


Question 8

(i) sin35cos55+cos35sin55cosec210tan280

(ii) sin234+sin256+2tan18tan72cot230

Sol :

Given :

(i) sin35cos55+cos35sin55cosec210tan280

=sin35cos(9035)+cos35sin(9035)cosec210tan2(9010)

=sin35sin35+cos35cos35cosec210cot210

=sin235+cos235cosec210cot210=11=1 {sin2θ+cos2θ=1cosec2θcot2θ=1}


(ii) sin234+sin256+2tan18tan72cot230

=sin234+sin2(9034)+2tan18tan(9018)cot230

=sin234+cos234+2tan18cot18cot230

=1+2×1(3)2

=1+2-3=0


Question 9

(i) (tan25cosec65)2+(cot25sec65)2+2tan18tan45tan72

(ii) (cos225+cos265)+cosecθsec(90θ)cotθtan(90θ)

Sol :

(i) (tan25cosec65)2+(cot25sec65)2+2tan18tan45tan72

=(tan25cosec(9025))2+(cot25sec(9025))2+2tan18tan(9018)tan45

=(tan25sec25)2+(cot25cosec25)2+2tan18cot18tan45

=(sin25×cos25cos25×1)2+(cos25×sin25sin25×1)+2×1×1

=sin225+cos225+2

=1+2=3 {sin2θ+cos2θ=1tanθcotθ=1}


(ii) (cos225+cos265)+cosecθsec(90θ)cotθtan(90θ)

=[cos225+cos2(9025)]+cosecθcosecθcotθcotθ

=(cos225+sin225)+(cosec2θcot2θ)

=1+1=2


Question 10

(i) 2(sec235cot255)cos28cosec62tan18tan36tan30tan54tan72

(ii) cosec2(90θ)tan2θ2(cos248+cos242)2tan230sec252sin238cosec270tan220

Sol :

(i) 2(sec235cot255)cos28cosec62tan18tan36tan30tan54tan72

=2[sec235cot2(9035)]cos28cosec(9028)tan18tan(9018)tan36tan(9036)tan30

=2(1)11×1×13=231=23 {sec2θtan2θ=1tanθcotθ=1cosθsecθ=1}


(ii) cosec2(90θ)tan2θ2(cos248+cos242)2(tan230sec252sin238)cosec270tan220

=sec2θtan2θ2(cos248+cos2(9048)2[tan230sec252sin2(9052)cosec270tan2(9070)

=12[cos248+sin248]2[(13)2sec252cos252]cosec270cot270

=12×12[13×1]1

=1223  {sin2θ+cos2θ=1sec2θtan2θ=1cosec2θcot2θ=1}

=346=16


Question 11

Prove that following:

(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1

(ii) tanθtan(90θ)+sin(90θ)cosθ=sec2θ

(iii) cos(90θ)cosθtanθ+cos2(90θ)=1

(iv) sin(90θ)cos(90θ)=tanθ1+tan2θ

Sol :

(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1

L.H.S. = cos θ sin (90° – θ) + sin θ cos (90° – θ)

= cos θ . cos θ + sin θ . sin θ

=cos2θ+sin2θ = 1 = R.H.S.


(ii) tanθtan(90θ)+sin(90θ)cosθ=sec2θ

L.H.S. =tanθtan(90θ)+sin(90θ)cosθ

=tanθcotθ+cosθcosθ=tanθ×tanθ+1

=tan2θ+1=sec2θ=R.H.S


(iii) cos(90θ)cosθtanθ+cos2(90θ)=1

L.H.S=cos(90θ)cosθtanθ+cos2(90θ)

=sinθcosθsinθcosθ+sin2θ

=sinθcosθ×cosθsinθ+sin2θ

=cos2θ+sin2θ=1=R.H.S


(iv) sin(90θ)cos(90θ)=tanθ1+tan2θ

L.H.S=sin(90θ)cos(90θ)

=cosθsinθ {sin(90θ)cosθsin2θ+cos2θ=1}

R.H.S=tanθ1+tan2θ=sinθcosθ1+sin2θcos2θ

=sinθcosθcos2θ+sin2θcos2θ

=sinθcosθ1cos2θ=sinθcosθ×cos2θ=sinθcosθ

∴L.H.S=R.H.S

Prove that following (12 to 30) identities, where the angles involved are acute angles for which the trigonometric ratios as defined:


Question 12

(i) (sec A + tan A) (1 – sin A) = cos A

(ii) (1 + tan2A) (1 – sin A) (1 + sin A) = 1.

Sol :

(i) (sec A + tan A) (1 – sin A) = cos A

L.H.S. = (sec A + tan A) (1 – sin A)

=(1cosA+sinAcosA)(1sinA)

=cos2AcosA=cosA=R.H.S

(1sin2A=cos2A)


(ii) (1+tan2A)(1sinA)(1+sinA)=1

L.H.S=(1+tan2A)(1sinA)(1+sinA)

=(1+sin2 Acos2 A)(1sin2 A)

=cos2A+sin2Acos2A×cos2A

{1sin2A=cos2Asin2A+cos2A=1}

=1cos2A×cos2A=1=R.H.S


Question 13

(i) tan A + cot A = sec A cosec A

(ii) (1 – cos A)(1 + sec A) = tan A sin A.

Sol :

(i) tan A + cot A = sec A cosec A

L.H.S. = tan A + cot A

=sinAcosA+cosAsinA

=sin2A+cos2AsinAcosA

=1sinAcosA=cosecAsecA=secAcosec

A=R.H.S


(ii) (1-cos A)(1+sec A)=tan A sin A

L.H.S=(1-cos A)(1+sec A)

=(1cosA)(1+1cosA)

=(1cosA)(cosA+1)cosA

=(1cosA)(1+cosA)cosA=1cos2AcosA

=sin2AcosA

=sin2AcosA=sinA×sinAcosA {1cos2A=sin2A}

=tan A sin A =R.H.S


Question 14

(i) 11+cosA+11cosA=2cosec2A

(ii) 1secA+tanA+1secAtanA=2secA

Sol :

(i) 11+cosA+11cosA=2cosec2A

L.H.S=11+cosA+11cosA

=1cosA+1+cosA(1+cosA)(1cosA) 

=21cos2A=2sin2A (1cos2A=sin2A)

=2cosec2A=R.H.S


(ii) 1secA+tanA+1secAtanA=2secA

L.H.S =1secA+tanA+1secAtanA

=secAtanA+secA+tanA(secA+tanA)(secAtanA)

=2secAsec2Atan2A=2secA1 (sec2Atan2A=1)

=2 sec A =R.H.S


Question 15

(i) sinA1+cosA=1cosAsinA

(ii) 1tan2Acot2A1=tan2A

(iii) sinA1+cosA=cosecAcotA

Sol :

(i) sinA1+cosA=1cosAsinA

L.H.S=sinA1+cosA

[multiplying and dividing by (1 – cosA)]

=sinA(1cosA)1cos2A=sinA(1cosA)sin2A

(1cos2A=sin2A)

=1cosAsinA=R.H.S


(ii) 1tan2Acot2A1=tan2A

L.H.S=1tan2Acot2A1

=1sin2Acos2Acos2Asin2A1

=cos2Asin2Acos2Acos2Asin2Asin2A

=cos2Asin2Acos2A×sin2Acos2Asin2A

=sin2Acos2A=tan2A= R.H.S


(iii) sinA1+cosA=cosecAcotA

R.H.S=cosec A- cot A

=1sinAcosAsinA=1cosAsinA

=(1cosA)(1+cosA)sinA(1+cosA)

( Multiplying and dividing by 1+cos A)

=1cos2AsinA(1+cosA)=sin2AsinA(1+cosA)

{1cos2A=sin2A}

sinA1+cosA=L.H .S


Question 16

(i) secA1secA+1=1cosA1+cosA

(ii) tan2θ(secθ1)2=1+cosθ1cosθ

(iii) (1+tanA)2+(1tanA)2=2sec2A

(iv) sec2A+cosec2A=sec2Acosec2A

Sol :

(i) secA1secA+1=1cosA1+cosA

L.H.S=secA1secA+1

=1cosAcosA1+cosAcosA=1cosAcosA×cosA1+cosA

=1cosA1+cosA=R.H.S


(ii) Prove that tan2θ(secθ1)2=1+cosθ1cosθ

L.H.S=tan2θ(secθ1)2=tan2θsec2θ+12secθ

=sin2θcos2θ1cos2θ+12cosθ

=sin2θcos2θ×cos2θ1+cos2θ2cosθ

=sin2θ(1cosθ)2

=(1cos2θ)(1cosθ)2

=(1+cosθ)(1cosθ)(1cosθ)2

=1+cosθ1cosθ

=R.H.S


(ii) (1+tanA)2+(1tanA)2=2sec2A

L.H.S =(1+tanA)2+(1tanA)2

=1+2tanA+tan2A+12tanA+tan2A

=2+2tan2A=2(1+tan2A)

=2sec2A (1+tan2A=sec2A)

=R.H.S


(iv) sec2A+cosec2A=sec2Acosec2A

L.H.S=sec2A+cosec2A

=1cos2A+1sin2A

=sin2A+cos2Asin2A+cos2A

=1sin2Acos2A=sec2Acosec2A

=R.H.S


Question 17

(i) 1+sinAcosA+cosA1+sinA=2secA

(ii) tanAsecA1+tanAsecA+1=2cosecA

Sol :

(i) 1+sinAcosA+cosA1+sinA=2secA

L.H.S=1+sinAcosA+cosA1+sinA

=(1+sinA)(1+sinA)+cos2AcosA(1+sinA)

=1+sinA+sinA+sin2A+cos2AcosA(1+sinA)

=1+2sinA+1cosA(1+sinA)=2+2sinAcosA(1+sinA)

=2(1+sinA)cosA(1+sinA)=2cosA=2secA

=R.H.S


(ii) tanAsecA1+tanAsecA+1=2cosecA

L.H.S=tanAsecA1+tanAsecA+1

=tanA(1secA1+1secA+1)

=tanA(secA+1+secA1(secA1)(secA+1))

=tanA×2secAsec2A1

=2secAtanAtan2A

=2secAtanA

=2×1×cosAcosA×sinA

=2sinA=2cosecA

=R.H.S


Question 18

(i) cosecAcosecA1+cosecAcosecA+1=2sec2A

(ii) cotAtanA=2cos2A1sinAcosA

(iii) cotA12sec2A=cotA1+tanA

Sol :

(i) cosecAcosecA1+cosecAcosecA+1=2sec2A

L.H.S=cosecAcosecA1+cosecAcosecA+1

=cosecA[1cosecA1+1cosecA+1]

=cosecA[cosecA+1+cosecA1(cosecA1)(cosecA+1)]

=cosecA×2cosecAcosec2A1=2cosec2Acot2A

=2×sin2Asin2A×cos2A=2cos2A

=2sec2A=R.H.S


(ii) cotAtanA=2cos2A1sinAcosA

L.H.S=cot A-tan A

=cosAsinAsinAcosA=cos2Asin2AsinAcosA

=cos2A(1cos2A)sinAcosA

=cos2A1+cos2AsinAcosA

=2cos2A1sinAcosA

=R.H.S


(iii) cotA12sec2A=cotA1+tanA

L.H.S=cotA12sec2A

=cosAsinA121cos2A

=cosAsinAsinA2cos2A1cos2A

=cosAsinAsinA×cos2A2cos2A1

=cos2A(cosAsinA)sinA(2cos2A1)

=cos2A(cosAsinA)sinA[2cos2A(sin2A+cos2A)]

=cos2A(cosAsinA)sinA[2cos2Asin2Acos2A]

=cos2A(cosAsinA)sinA(cos2Asin2A)

=cos2A(cosAsinA)sinA(cosA+sinA)(cosAsinA)

=cos2AsinA(cosA+sinA)


R.H.S=cotA1+tanA=cosAsinA1+sinAcosA

=cosAsinAcosA+sinAcosA

=cosAsinA×cosAcosA+sinA

=cos2AsinA(cosA+sinA)

∴L.H.S=R.H.S


Question 19

(i) tan2θsin2θ=tan2θsin2θ

(ii) cosθ1tanθsin2θcosθsinθ=cosθ+sinθ

Sol :

(i) tan2θsin2θ=tan2θsin2θ

L.H.S =tan2θsin2θ

=sin2θcos2θsin2θ

=sin2θsin2θcos2θcos2θ

=sin2θ(1cos2θ)cos2θ=sin2θ×sin2θcos2θ

=sin2θ×tan2θ

=tan2θsin2θ

=R.H.S


(ii) cosθ1tanθsin2θcosθsinθ=cosθ+sinθ

L.H.S=cosθ1tanθsin2θcosθsinθ

=cosθ1sinθcosθsin2θcosθsinθ

=cosθcosθsinθcosθsin2θcosθsinθ

=cos2θcosθsinθsin2θcosθsinθ

=cos2θsin2θcosθsinθ

=(cosθ+sinθ)(cosθsinθ)cosθsinθ

=cos θ+sin θ=R.H.S


Question 20

(i) cosec4θcosec2θ=cot4θ+cot2θ

(ii) 2sec2θsec4θ2cosec2θ+cosec4θ=cot4θtan4θ

Sol :

(i) cosec4θcosec2θ=cot4θ+cot2θ

L.H .S=cosec4θcosec2θ

=cosec2θ(cosec2θ1)

=cosec2θcot2θ (cosec2θ1=cot2θ)

=(cot2θ+1)cot2θ

=cot4θ+cot2θ

=R.H.S


(ii) 2sec2θsec4θ2cosec2θ+cosec4θ=cot4θtan4θ

L.H.S=2sec2θsec4θ2cosec2θ+cosec4θ

=2(tan2θ+1)(tan2θ+1)22(1+cot2θ)+(1+cot2θ)2

{sec2θ=tan2θ+1cosec2θ=1+cot2θ}

=2tan2θ+2(tan4θ+2tan2θ+1)22cot2θ+(1+2cot2θ+cot4θ)

=2tan2θ+2tan4θ2tan2θ122cot2θ+1+2cot2θ+cot4θ

=cot4θtan4θ=R.H.S


Question 21

(i) 1+cosθsin2θsinθ(1+cosθ)=cotθ

(ii) tan3θ1tanθ1=sec2θ+tanθ

Sol :

(i) 1+cosθsin2θsinθ(1+cosθ)=cotθ

L.H.S=1+cosθsin2θsinθ(1+cosθ)

L.H.S. =1+cosθsin2θsinθ(1+cosθ)

=cosθ+cos2θsinθ(1+cosθ)

=cosθ(1+cosθ)sinθ(1+cosθ)

=cosθsinθ=cotθ

=R.H.S


(iii) tan3θ1tanθ1=sec2θ+tanθ

L.H.S. =(tanθ1)tanθ1(tan2θ+tanθ+1)

=tan2θ+tanθ+1=tan2θ+1+tanθ

=sec2θ+tanθ

=R.H.S


Question 22

(i) 1+cosecAcosecA=cos2A1sinA

(ii) 1cosA1+cosA=sinA1+cosA

Sol :

(i) 1+cosecAcosecA=cos2A1sinA

L.H.S=1+cosecAcosecA

=1+cosecAcosecA

=1+1sinA1sinA

=sinA+1sinA×sinA1

=sin A+1


R.H.S=cos2A1sinA=1sin2A1sinA

=(1+sinA)(1sinA)1sinA

=1+sin A=sin A+1

∴L.H.S=R.H.S


(ii) 1cosA1+cosA=cosecAcotA

L.H.S.=1cosA1+cosA

Rationalising the denominator

=(1cosA)(1cosA)(1+cosA)(1cosA)

=(1cosA)21cos2A

=(1cosA)2sin2A

=1cosAsinA

=1sinAcosAsinA

=cosec A-cot A=R.H.S


Question 23

(i) 1+sinA1sinA=tanA+secA

(ii) 1cosA1+cosA=cosecAcotA

Sol :

(i) 1+sinA1sinA=tanA+secA

L.H.S. =1+sinA1sinA

Rationalising the denominator

=(1+sinA)(1+sinA)(1sinA)(1+sinA)

=(1+sinA)21sin2 A

=(1+sinA)2cos2 A

=1+sinAcosA

=1cosA+sinAcosA

=sec A+ tan A

=tan A+ sec A=R.H.S


(ii) 1cosA1+cosA=cosecAcotA

L.H.S. =1cosA1+cosA

Rationalising the denominator

=(1cosA)(1cosA)(1+cosA)(1cosA)

=(1cosA)21cos2A

=(1cosA)2sin2A

=1cosAsinA

=1sinAcosAsinA

=cosec A-cot A=R.H.S


Question 24

(i) secA1secA+1+secA+1secA1=2cosecA

(ii) cotAcotA1sinA=1+cosecA

Sol :

(i) secA1secA+1+secA+1secA1=2cosecA

LH.S=secA1secA+1+secA+1secA1

=secA1secA+1+secA+1secA1

=secA1+secA+1(secA+1)(secA1)

=2secAsec2A1

{sec2A1=tan2A}

=2secAtan2A=2secAtanA

=2×cosAcosA×sinA=2sinA

=2cosecA=R.H.S .


(ii) cosAcotA1sinA=1+cosecA

L.H.S =cosAcotA1sinA=cosAcosAsinA(1sinA)

{cosA=cosAsinA}

=cos2AsinA(1sinA)=1sin2AsinA(1sinA)

{cos2A=1sin2A}

=(1+sinA)(1sinA)sinA(1sinA)=1+sinAsinA

=1sinA+sinAsinA=cosecA+1

=1+cosec A=R.H.S


Question 25

(i) 1+tanAsinA+1+cotAcosA=2(secA+cosecA)

(ii) sec4Atan4A=1+2tan2A

Sol :

(i) 1+tanAsinA+1+cotAcosA=2(secA+cosecA)

L.H.S=1+tanAsinA+1+cotAcosA

=1+sinAcosAsinA+1+cosAsinAcosA

=cosA+sinAcosA×sinA+sinA+cosAcosA×sinA

=2[cosA+sinAcosAsinA]

=2[cosAcosAsinA+sinAcosAsinA]

=2[1sinA+1cosA]

=2(cosec A+ sec A)

=2(sec A+cosec A)

=R.H.S


(ii) sec4Atan4A=1+2tan2A

L.H.S=sec4Atan4A

=(sec2Atan2A)(sec2A+tan2A)

=(1+tan2Atan2A)(1+tan2A+tan2A)

{sec2A=tan2A+1}

=1(1+2tan2A)=1+2tan2A=R.H.S


Question 26

(i) cosec6Acot6A=3cot2Acosec2A+1

(ii) sec6Atan6A=1+3tan2A+3tan4A

Sol :

(i) cosec6Acot6A=3cot2Acosec2A+1

L.H.S=cosec6Acot6A

=(cosec2A)3(cot2A)3

=(cosec2θcot2A)3+3cosec2Acot2A(cosec2Acot2A)

=(1)3+3cosec2Acot2A×1

=1+3cot2Acosec2A

=3cot2Acosec2A+1=R.H.S


(ii) sec6Atan6A=1+3tan2A+3tan4A

L.H.S=sec6Atan6A

=(sec2A)3(tan2A)3

=(sec2Atan2A)3+3sec2Atan2A(sec2Atan2A)

=(1)3+3sec2 Atan2 A×1

=1+3sec2 Atan2 A

=1+3[(1+tan2A)(tan2A)]

=1+3[tan2A+tan4A]

=1+3tan2A+3tan4A=R.H.S


Question 27

(i) cotθcosecθ1cotθcosecθ+1=1+cosθsinθ

(ii) sinθcotθ+cosecθ=2+sinθcotθcosecθ

Sol :

(i) cotθcosecθ1cotθcosecθ+1=1+cosθsinθ

L.H.S =cotθcosecθ1cotθcosecθ+1

=cosθsinθ+1sinθ1cosθsinθ1sinθ+1

=cosθ+1sinθsinθ×sinθcosθ1+sinθ

=cosθ+1sinθcosθ1+sinθ

=cosθ+(1sinθ)cosθ(1sinθ)

=[cosθ+(1sinθ)][cosθ+(1sinθ)][cosθ(1sinθ)][cosθ+(1sinθ)]

=[cosθ+(1sinθ)]2cos2θ(1sinθ)2

=(cosθ+1sinθ)2cos2θ(1+sin2θ2sinθ)

=cos2θ+sin2θ+1+2cosθ2sinθ2sinθcosθcos2θ1sin2θ+2sinθ

=1+1+2cosθ2sinθ2sinθcosθ1sin2θ1sin2θ+2sinθ

=2+2cosθ2sinθ2sinθcosθ2sinθ2sin2θ

=2(1+cosθ)2sinθ(1+cosθ)2sinθ(1sinθ)

=(1+cosθ)2(1sinθ)2sinθ(1sinθ)

=1+cosθsinθ=R.H.S


(ii) sinθcotθ+cosecθ=2+sinθcotθcosecθ

=sin2θ1+cosθ

=1cos2θ1+cosθ=(1+cosθ)(1cosθ)1+cosθ

=1-cos θ


R.H.S=2+sinθcotθcosecθ

=2+sinθcosθsinθ1sinθ

=2+sinθcosθ1sinθ

=2+sin2θcosθ1

=2cosθ2+sin2θcosθ1

=2cosθ2+(1cos2θ)cosθ1

=2(cosθ1)+(1+cosθ)(1cosθ)cosθ1


Question 28

(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ

(ii) (cosecA – sinA)(secA – cosA) sec2A = tanA

(iii) (cosecθ – sinθ)(secθ – cosθ)(tan θ + cotθ) = 1

Sol :
(i) (sinθ + cosθ)(secθ + cosecθ) = 2 + secθ cosecθ
L.H.S = (sinθ + cosθ)(secθ + cosecθ)
=(sinθ+cosθ)(1cosθ+1sinθ)

=(sinθ+cosθ)(sinθ+cosθ)sinθcosθ

=sin2θ+sinθcosθ+sinθcosθ+cos2θsinθcosθ

=1+2sinθcosθsinθcosθ

=1sinθcosθ+2sinθcosθsinθcosθ

=cosec ፀ sec ፀ +2

=2+secθ cosecθ


(ii) (cosec A-sin A)(sec A-cos A)sec2A

=tan A

L.H.S

(cosecAsinA)(secAcosA)sec2A

=(1sinAsinA)(IcosAcosA)1cos2 A

=(1sin2AsinA)(1cos2AcosA)1cos2A

cos2AsinAsin2AcosA1cos2A=sinAcosA=tanA

R.H.S


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

L.H.S=(cosecθ-sinθ)(secθ-cosθ)(tanθ+cotθ)

=(1sinθsinθ)(1cosθcosθ)(tanθ+cotθ)

=1sin2θsinθ×1cos2θcosθ(tanθ+cotθ)

=cos2θsinθ×sin2θcosθ(tanθ+cotθ)

=sinθcosθsin2θ+cos2θsinθcosθ

=1=R.H.S

Question 29

(i) sin3A+cos3AsinA+cosA+sin3Acos3AsinAcosA=2

(ii) tan2A1+tan2A+cot2A1+cot2A=1

Sol :

(i) sin3A+cos3AsinA+cosA+sin3Acos3AsinAcosA=2

L.H.S=sin3A+cos3AsinA+cosA+sin3Acos3AsinAcosA

=(sinA+cosA)(sin2AsinAcosA+cos2A)(sinA+cosA)+(sinAcosA)(sin2A+sinAcosA+cos2A)(sinAcosA)

=(1sinAcosA)+(1+sinAcosA)  [sin2A+cos2A=1}

=1sinAcosA+1+sinAcosA=2=R.H.S


(ii) tan2A1+tan2A+cot2A1+cot2A=1

=tan2A1+tan2A+1tan2A1+1tan2A

=tan2A1+tan2A+1tan2Atan2A+1tan2A

=tan2A1+tan2A+tan2Atan2A(tan2A+1)

=tan2A1+tan2A+11+tan2A

=1+tan2A1+tan2A=1

=R.H.S

Question 30

(i) 1secA+tanA1cosA=1cosA1secAtanA

(ii) (sinA+secA)2+(cosA+cosecA)2=(1+secAcosecA)2

(iii) tanA+sinAtanAsinA=secA+1secA1

Sol :

(i) 1secA+tanA1cosA=1cosA1secAtanA

L.H.S=1secA+tanA1cosA

=11cosA+sinAcosA1cosA

=cosA1+sinA1cosA

=cos2A1sinAcosA(1+sinA)=sin2sinAcosA(1+sinA)

=sinA(1+sinA)cosA(1+sinA)=tanA


R.H.S=\frac{1}{\cos A}-\frac{1}{\sec A-\tan A}

=\frac{1}{\cos A}-\frac{1}{\frac{1}{\cos A}-\frac{\sin A}{\cos A}}

=\frac{1}{\cos A}-\frac{\cos A}{1-\sin A}

=\frac{1-\sin A-\cos ^{2} A}{\cos A(1-\sin A)}

=\frac{\sin ^{2} A-\sin A}{\cos A(1-\sin A)}

=\frac{-\sin A+\sin ^{2} A}{\cos A(1-\sin A)}

=\frac{-\sin A(1-\sin A)}{\cos A(1-\sin A)}

\frac{-\sin A}{\cos A}=-\tan A

∴L.H.S=R.H.S


(ii) (\sin A+\sec A)^{2}+(\cos A+\operatorname{cosec} A)^{2}=(1+\sec A \operatorname{cosec} A)^{2}

L.H.S=(\sin A+\sec A)^{2}+(\cos A+\operatorname{cosec} A)^{2}

=\sin ^{2} A+\sec ^{2} A+2 \sin A \sec A+\cos ^{2} A+\operatorname{cosec}^{2} A+2 \cos A \operatorname{cosec} A

=\left(\sin ^{2} A+\cos ^{2} A\right)+\left(\sec ^{2} A+\operatorname{cosec}^{2} A\right)+2 \sin A \times \frac{1}{\cos A}+2 \times \cos A \times \frac{1}{\sin A}

=1+\left[\frac{1}{\cos ^{2} A}+\frac{1}{\sin ^{2} A}\right]+\frac{2 \sin ^{2} A+2 \cos ^{2} A}{\sin A \cos A}

=1+\left[\frac{\sin ^{2} A+\cos ^{2} A}{\cos ^{2} A \sin ^{2} A}\right]+\frac{2\left[\sin ^{2} A+\cos ^{2} A\right]}{\sin A \cos A}

=1+\frac{1}{\cos ^{2} A \sin ^{2} A}+\frac{2}{\sin A \cos A}

\left[\because \sin ^{2} \theta+\cos ^{2} \theta+1\right]

=\left(1+\frac{1}{\cos A \sin A}\right)^{2} \left.\left[\because(a+b)^{2}=a^{2}+(b)^{2}+2 a b\right)\right]

=(1+\operatorname{cosec} A \sec A)^{2}

=R.H.S


(iii) \frac{\tan A+\sin A}{\tan A-\sin A}=\frac{\sec A+1}{\sec A-1}

L.H.S=\frac{\tan A+\sin A}{\tan A-\sin A}

=\frac{\frac{\sin A}{\cos A}+\sin A}{\frac{\sin A}{\cos A}-\sin A}

=\frac{\frac{\sin A+\sin A \cos A}{\cos A}}{\frac{\sin A-\sin A \cos A}{\cos A}}

=\frac{\sin A(1+\cos A)}{\sin A(1-\cos A)}=\frac{1+\cos A}{1-\cos A}

Dividing each term by cos A

\frac{\frac{1}{\cos A}+1}{\frac{1}{\cos A}-1}=\frac{\sec A+1}{\sec A-1}=R.H.S


Question 31

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

Sol :
sin θ + cos θ = √2 sin (90° – θ)
sin θ + cos θ = √2 cos θ
dividing by sin θ
1+\cot \theta=\sqrt{2} \cot \theta
1=\sqrt{2} \cot \theta-\cot \theta
1=(\sqrt{2}-1) \cot \theta

\cot \theta=\frac{1}{\sqrt{2}-1}=\frac{1 \times(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} (rationalising the denominator)

=\frac{(\sqrt{2}+1)}{(\sqrt{2})^{2}-(1)^{2}}=\frac{\sqrt{2}+1}{2-1}=\frac{\sqrt{2}+1}{1}

=\sqrt{2}+1

=R.H.S


Question 32

If 7 \sin ^{2} \theta+3 \cos ^{2} \theta=4,0^{\circ} \leq \theta \leq 90^{\circ}, then find the value of θ.

Sol :
7 \sin ^{2} \theta+3 \cos ^{2} \theta=4,0^{\circ} \leq \theta \leq 90^{\circ}
3 \sin ^{2} \theta+3 \cos ^{2} \theta+4 \sin ^{2} \theta=4

3\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=4-4 \sin ^{2} \theta

3 \times 1=4\left(1-\sin ^{2} \theta\right) \Rightarrow \frac{3}{4}=\cos ^{2} \theta

\cos \theta=\frac{\sqrt{3}}{2}=\cos 30^{\circ}

\therefore \theta=30^{\circ}


Question 33

If sec θ + tan θ = m and sec θ – tan θ = n, prove that mn = 1.

Sol :

sec θ + tan θ = m and sec θ – tan θ = n

mn = (sec θ + tan θ) (sec θ – tan θ) 
=\sec ^{2} \theta-\tan ^{2} \theta=1 

\left(\therefore \sec ^{2} \theta-\tan ^{2} \theta=1\right)

Hence proved.


Question 34

If x – a sec θ + b tan θ and y = a tan θ + b sec θ, prove that x^{2}-y^{2}=a^{2}-b^{2}

Sol :
x – a sec θ + b tan θ and y = a tan θ + b sec θ
To prove that x^{2}-y^{2}=a^{2}-b^{2}

=\left(a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta+2 a b \sec \theta \tan \theta\right)-\left(a^{2} \tan ^{2} \theta+b^{2} \sec ^{2} \theta+2 a b \sec \theta \tan \theta\right)

=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta+2 a b \sec \theta \tan \theta-a^{2} \tan ^{2} \theta-b^{2} \sec ^{2} \theta-2 a b \sec \theta \tan \theta

=a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)-b^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)

=a^{2} \times 1-b^{2} \times 1 \left\{\sec ^{2} \theta-\tan ^{2} \theta=1\right\}

=a^{2}-b^{2}

Hence proved


Question 35

If x = h + a cos θ and y = k + a sin θ, prove that (x-h)^{2}+(y-k)^{2}=a^{2}

Sol :

x = h + a cos θ and y = k + a sin θ

To prove that (x-h)^{2}+(y-k)^{2}=a^{2}

(x-h)=a \cos \theta
(y-k)=a \sin \theta

Squaring and adding,

(x-h)^{2}+(y-k)^{2}=a^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta

=a^{2} \cdot\left(\sin ^{2} \theta+\cos ^{2} \theta\right) \left\{\because \sin ^{2} \theta+\cos ^{2} \theta=1\right)

=a^{2} \times 1=a^{2}

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