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).
In ∆ABC, ∠B = 90°
AC = 5 and BC = 3
=√25−9=√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).
In right ∆ABC
AC = 17, AB = 8
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.
cosA=√1−sin2A
tanA=sinAcosA=sinA√1−sin2A
secA=1cosA=1√1−sin2A
Question 4
If tanA=1√3, find all other trigonometric ratios of angle A.
In right ∆ABC,
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=2√3
cosecA=1sinA=21=2
Question 5
If 12 cosec θ = 13, find the value of 2sinθ−3cosθ4sinθ−9cosθ
Sol :
12 cosec θ = 13
In right ∆ABC,
∠A = θ
Figure to be added
∴AC=13,BC=12
AB=√AC2−BC2=√132−122
=√169−144=√25=5
Now sinθ=BCAC=1213
cosθ=ABAC=513
Now 2sinθ−3cosθ4sinθ−9cosθ=2×1213−3×5134×1213−9×513
=2413−15134813−4513
=913313=913×133=3
Without using trigonometric tables, evaluate the following (6 to 10) :
Question 6
(i) cos226∘+cos64∘sin26∘+tan36∘cot54∘
(ii) sec17∘cosec73∘+tan68∘cot22∘+cos244∘+cos246∘
Sol :
Given that
=cos2⋅26∘+sin226∘+tan36∘tan36∘
=1+1=2
[∵cos(90∘−θ)=sinθ and cot(90∘−θ)=tanθ,sin2θ+cos2θ=1]
(ii) sec17∘cosec73∘+tan68∘cot22∘+cos244∘+cos246∘
sec17∘cosec73∘+tan68∘cot22∘+cos244∘+cos246∘
=sec(90∘−73∘)cosec(73∘)+tan(90∘−22∘)cot22∘+cos2(90∘−46∘+cos246∘
=cosec73∘cosec73∘+cot22∘cot22∘+sin246∘+cos246∘ (∵sin2θ+cos2θ=1)
=1+1+1=3
Question 7
(i) sin65∘cos25∘+cos32∘sin58∘−sin28∘sec62∘+cosec230∘(2015)
(ii) sin29∘cosec61∘+2cot8∘cot17∘cot45∘cot73∘cot82∘−3(sin238∘+sin252∘)
Sol :
Given :
(i) sin65∘cos25∘+cos32∘sin58∘−sin28∘sec62∘+cosec230∘
=sin65∘cos(90∘−65∘)+cos32∘sin(90∘−32∘)−sin28∘×sec(90∘−28∘)+(2)2
=sin65∘sin65∘+cos32∘cos32∘−sin28∘×cosec28∘+4 {sin(90∘+θ)=cosθcos(90∘−θ)=sinθ}
=1+1-1+4
=6-1=5
(ii) sin29∘cosec61∘+2cot8∘cot17∘cot45∘cot73∘cot82∘−3(sin238∘+sin252∘)
=sec29∘cosec(90∘−29∘)+2cot8∘cot82∘cot17∘cot73∘cot45∘−3(sin238∘+sin2(90∘−38∘)
sec29∘sec29∘+2cot8cot(90∘−8∘)cot17∘cot(90∘−17∘)cot45∘−3(sin238∘+cos238∘)
=1+2cot8∘tan8∘cot17∘cot17∘cot45∘−3(1)
{sec(90∘−θ)=cosecθ,cotθ=tan(90∘−θ) and sin2θ+cos2θ=1}
=1+2×1×1×1−3(cot45∘=1)
=1+2-3=0
Question 8
(i) sin35∘cos55∘+cos35∘sin55∘cosec210∘−tan280∘
(ii) sin234∘+sin256∘+2tan18∘tan72∘−cot230∘
Sol :
Given :
(i) sin35∘cos55∘+cos35∘sin55∘cosec210∘−tan280∘
=sin35∘cos(90∘−35∘)+cos35∘sin(90∘−35∘)cosec210∘−tan2(90∘−10∘)
=sin35∘sin35∘+cos35∘cos35∘cosec210∘−cot210∘
=sin235∘+cos235∘cosec210∘−cot210∘=11=1 {∵sin2θ+cos2θ=1cosec2θ−cot2θ=1}
(ii) sin234∘+sin256∘+2tan18∘tan72∘−cot230∘
=sin234∘+sin2(90∘−34∘)+2tan18∘tan(90∘−18∘)−cot230∘
=sin234∘+cos234∘+2tan18∘cot18∘−cot230∘
=1+2×1−(√3)2
=1+2-3=0
Question 9
(i) (tan25∘cosec65∘)2+(cot25∘sec65∘)2+2tan18∘tan45∘tan72∘
(ii) (cos225+cos265)+cosecθsec(90∘−θ)−cotθtan(90∘−θ)
Sol :
(i) (tan25∘cosec65∘)2+(cot25∘sec65∘)2+2tan18∘tan45∘tan72∘
=(tan25∘cosec(90∘−25∘))2+(cot25∘sec(90∘−25∘))2+2tan18∘tan(90∘−18∘)tan45∘
=(tan25∘sec25∘)2+(cot25∘cosec25∘)2+2tan18∘cot18∘tan45∘
=(sin25∘×cos25∘cos25∘×1)2+(cos25∘×sin25∘sin25∘×1)+2×1×1
=sin225∘+cos225∘+2
=1+2=3 {∵sin2θ+cos2θ=1tanθcotθ=1}
(ii) (cos225∘+cos265∘)+cosecθsec(90∘−θ)−cotθtan(90∘−θ)
=[cos225∘+cos2(90∘−25∘)]+cosecθcosecθ−cotθ⋅cotθ
=(cos225∘+sin225∘)+(cosec2θ−cot2θ)
=1+1=2
Question 10
(i) 2(sec235∘−cot255∘)−cos28∘cosec62∘tan18∘tan36∘tan30∘tan54∘tan72∘
(ii) cosec2(90−θ)−tan2θ2(cos248∘+cos242∘)−2tan230∘sec252∘sin238∘cosec270∘−tan220∘
Sol :
(i) 2(sec235∘−cot255∘)−cos28∘cosec62∘tan18∘tan36∘tan30∘tan54∘tan72∘
=2[sec235∘−cot2(90∘−35∘)]−cos28∘cosec(90∘−28∘)tan18∘tan(90∘−18∘)tan36∘tan(90∘−36∘)tan30∘
=2(1)−11×1×1√3=2−√31=2−√3 {∵sec2θ−tan2θ=1tanθcotθ=1cosθsecθ=1}
(ii) cosec2(90∘−θ)−tan2θ2(cos248∘+cos242∘)−2(tan230∘sec252∘sin238∘)cosec270∘−tan220∘
=sec2θ−tan2θ2(cos248∘+cos2(90∘−48∘)−2[tan230∘sec252∘sin2(90∘−52∘)cosec270∘−tan2(90∘−70∘)
=12[cos248∘+sin248∘]−2[(1√3)2sec252∘cos252∘]cosec270∘−cot270
=12×1−2[13×1]1
=12−23 {∵sin2θ+cos2θ=1sec2θ−tan2θ=1cosec2θ−cot2θ=1}
=3−46=−16
Question 11
Prove that following:
(i) cos θ sin (90° – θ) + sin θ cos (90° – θ) = 1
(iii) cos(90∘−θ)cosθtanθ+cos2(90∘−θ)=1
(iv) sin(90∘−θ)cos(90∘−θ)=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.
=tanθcotθ+cosθcosθ=tanθ×tanθ+1
=tan2θ+1=sec2θ=R.H.S
(iii) cos(90∘−θ)cosθtanθ+cos2(90∘−θ)=1
L.H.S=cos(90∘−θ)cosθtanθ+cos2(90∘−θ)
=sinθcosθsinθcosθ+sin2θ
=sinθcosθ×cosθsinθ+sin2θ
=cos2θ+sin2θ=1=R.H.S
(iv) sin(90∘−θ)cos(90∘−θ)=tanθ1+tan2θ
L.H.S=sin(90∘−θ)cos(90∘−θ)
=cosθsinθ {∵sin(90∘−θ)−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)
=cos2AcosA=cosA=R.H.S
(1−sin2A=cos2A)
(ii) (1+tan2A)(1−sinA)(1+sinA)=1
L.H.S=(1+tan2A)(1−sinA)(1+sinA)
=(1+sin2 Acos2 A)(1−sin2 A)
=cos2A+sin2Acos2A×cos2A
{∵1−sin2A=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
=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)
=(1−cosA)(1+1cosA)
=(1−cosA)(cosA+1)cosA
=(1−cosA)(1+cosA)cosA=1−cos2AcosA
=sin2AcosA
=sin2AcosA=sinA×sinAcosA {1−cos2A=sin2A}
=tan A sin A =R.H.S
Question 14
(i) 11+cosA+11−cosA=2cosec2A
(ii) 1secA+tanA+1secA−tanA=2secA
Sol :
(i) 11+cosA+11−cosA=2cosec2A
L.H.S=11+cosA+11−cosA
=1−cosA+1+cosA(1+cosA)(1−cosA)
=21−cos2A=2sin2A (∵1−cos2A=sin2A)
=2cosec2A=R.H.S
(ii) 1secA+tanA+1secA−tanA=2secA
L.H.S =1secA+tanA+1secA−tanA
=secA−tanA+secA+tanA(secA+tanA)(secA−tanA)
=2secAsec2A−tan2A=2secA1 (∵sec2A−tan2A=1)
=2 sec A =R.H.S
Question 15
(i) sinA1+cosA=1−cosAsinA
(ii) 1−tan2Acot2A−1=tan2A
(iii) sinA1+cosA=cosecA−cotA
Sol :
(i) sinA1+cosA=1−cosAsinA
L.H.S=sinA1+cosA
[multiplying and dividing by (1 – cosA)]
(∵1−cos2A=sin2A)
=1−cosAsinA=R.H.S
(ii) 1−tan2Acot2A−1=tan2A
L.H.S=1−tan2Acot2A−1
=1−sin2Acos2Acos2Asin2A−1
=cos2A−sin2Acos2Acos2A−sin2Asin2A
=cos2A−sin2Acos2A×sin2Acos2A−sin2A
=sin2Acos2A=tan2A= R.H.S
(iii) sinA1+cosA=cosecA−cotA
R.H.S=cosec A- cot A
=1sinA−cosAsinA=1−cosAsinA
=(1−cosA)(1+cosA)sinA(1+cosA)
( Multiplying and dividing by 1+cos A)
=1−cos2AsinA(1+cosA)=sin2AsinA(1+cosA)
{∵1−cos2A=sin2A}
sinA1+cosA=L.H .S
Question 16
(i) secA−1secA+1=1−cosA1+cosA
(ii) tan2θ(secθ−1)2=1+cosθ1−cosθ
(iii) (1+tanA)2+(1−tanA)2=2sec2A
(iv) sec2A+cosec2A=sec2A⋅cosec2A
Sol :
(i) secA−1secA+1=1−cosA1+cosA
L.H.S=secA−1secA+1
=1−cosAcosA1+cosAcosA=1−cosAcosA×cosA1+cosA
=1−cosA1+cosA=R.H.S
(ii) Prove that tan2θ(secθ−1)2=1+cosθ1−cosθ
L.H.S=tan2θ(secθ−1)2=tan2θsec2θ+1−2secθ
=sin2θcos2θ1cos2θ+1−2cosθ
=sin2θcos2θ×cos2θ1+cos2θ−2cosθ
=sin2θ(1−cosθ)2
=(1−cos2θ)(1−cosθ)2
=(1+cosθ)(1−cosθ)(1−cosθ)2
=1+cosθ1−cosθ
=R.H.S
(ii) (1+tanA)2+(1−tanA)2=2sec2A
L.H.S =(1+tanA)2+(1−tanA)2
=1+2tanA+tan2A+1−2tanA+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) tanAsecA−1+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) tanAsecA−1+tanAsecA+1=2cosecA
L.H.S=tanAsecA−1+tanAsecA+1
=tanA(1secA−1+1secA+1)
=tanA(secA+1+secA−1(secA−1)(secA+1))
=tanA×2secAsec2A−1
=2secAtanAtan2A
=2secAtanA
=2×1×cosAcosA×sinA
=2sinA=2cosecA
=R.H.S
Question 18
(i) cosecAcosecA−1+cosecAcosecA+1=2sec2A
(ii) cotA−tanA=2cos2A−1sinA−cosA
(iii) cotA−12−sec2A=cotA1+tanA
Sol :
(i) cosecAcosecA−1+cosecAcosecA+1=2sec2A
L.H.S=cosecAcosecA−1+cosecAcosecA+1
=cosecA[1cosecA−1+1cosecA+1]
=cosecA[cosecA+1+cosecA−1(cosecA−1)(cosecA+1)]
=cosecA×2cosecAcosec2A−1=2cosec2Acot2A
=2×sin2Asin2A×cos2A=2cos2A
=2sec2A=R.H.S
(ii) cotA−tanA=2cos2A−1sinA−cosA
L.H.S=cot A-tan A
=cosAsinA−sinAcosA=cos2A−sin2AsinAcosA
=cos2A−(1−cos2A)sinAcosA
=cos2A−1+cos2AsinAcosA
=2cos2A−1sinAcosA
=R.H.S
(iii) cotA−12−sec2A=cotA1+tanA
L.H.S=cotA−12−sec2A
=cosAsinA−12−1cos2A
=cosA−sinAsinA2cos2A−1cos2A
=cosA−sinAsinA×cos2A2cos2A−1
=cos2A(cosA−sinA)sinA(2cos2A−1)
=cos2A(cosA−sinA)sinA[2cos2A−(sin2A+cos2A)]
=cos2A(cosA−sinA)sinA[2cos2A−sin2A−cos2A]
=cos2A(cosA−sinA)sinA(cos2A−sin2A)
=cos2A(cosA−sinA)sinA(cosA+sinA)(cosA−sinA)
=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) tan2θ−sin2θ=tan2θsin2θ
(ii) cosθ1−tanθ−sin2θcosθ−sinθ=cosθ+sinθ
Sol :
(i) tan2θ−sin2θ=tan2θsin2θ
L.H.S =tan2θ−sin2θ
=sin2θcos2θ−sin2θ
=sin2θ−sin2θcos2θcos2θ
=sin2θ(1−cos2θ)cos2θ=sin2θ×sin2θcos2θ
=sin2θ×tan2θ
=tan2θsin2θ
=R.H.S
(ii) cosθ1−tanθ−sin2θcosθ−sinθ=cosθ+sinθ
L.H.S=cosθ1−tanθ−sin2θcosθ−sinθ
=cosθ1−sinθcosθ−sin2θcosθ−sinθ
=cosθcosθ−sinθcosθ−sin2θcosθ−sinθ
=cos2θcosθ−sinθ−sin2θcosθ−sinθ
=cos2θ−sin2θcosθ−sinθ
=(cosθ+sinθ)(cosθ−sinθ)cosθ−sinθ
=cos θ+sin θ=R.H.S
Question 20
(i) cosec4θ−cosec2θ=cot4θ+cot2θ
(ii) 2sec2θ−sec4θ−2cosec2θ+cosec4θ=cot4θ−tan4θ
Sol :
(i) cosec4θ−cosec2θ=cot4θ+cot2θ
L.H .S=cosec4θ−cosec2θ
=cosec2θ(cosec2θ−1)
=cosec2θcot2θ (cosec2θ−1=cot2θ)
=(cot2θ+1)cot2θ
=cot4θ+cot2θ
=R.H.S
(ii) 2sec2θ−sec4θ−2cosec2θ+cosec4θ=cot4θ−tan4θ
L.H.S=2sec2θ−sec4θ−2cosec2θ+cosec4θ
=2(tan2θ+1)−(tan2θ+1)2−2(1+cot2θ)+(1+cot2θ)2
{∵sec2θ=tan2θ+1cosec2θ=1+cot2θ}
=2tan2θ+2−(tan4θ+2tan2θ+1)−2−2cot2θ+(1+2cot2θ+cot4θ)
=2tan2θ+2−tan4θ−2tan2θ−1−2−2cot2θ+1+2cot2θ+cot4θ
=cot4θ−tan4θ=R.H.S
Question 21
(i) 1+cosθ−sin2θsinθ(1+cosθ)=cotθ
(ii) tan3θ−1tanθ−1=sec2θ+tanθ
Sol :
(i) 1+cosθ−sin2θsinθ(1+cosθ)=cotθ
L.H.S=1+cosθ−sin2θsinθ(1+cosθ)
L.H.S. =1+cosθ−sin2θsinθ(1+cosθ)
=cosθ+cos2θsinθ(1+cosθ)
=cosθ(1+cosθ)sinθ(1+cosθ)
=cosθsinθ=cotθ
=R.H.S
(iii) tan3θ−1tanθ−1=sec2θ+tanθ
L.H.S. =(tanθ−1)tanθ−1(tan2θ+tanθ+1)
=tan2θ+tanθ+1=tan2θ+1+tanθ
=sec2θ+tanθ
=R.H.S
Question 22
(i) 1+cosecAcosecA=cos2A1−sinA
(ii) √1−cosA1+cosA=sinA1+cosA
Sol :
(i) 1+cosecAcosecA=cos2A1−sinA
L.H.S=1+cosecAcosecA
=1+cosecAcosecA
=1+1sinA1sinA
=sinA+1sinA×sinA1
=sin A+1
R.H.S=cos2A1−sinA=1−sin2A1−sinA
=(1+sinA)(1−sinA)1−sinA
=1+sin A=sin A+1
∴L.H.S=R.H.S
(ii) √1−cosA1+cosA=cosecA−cotA
L.H.S.=√1−cosA1+cosA
Rationalising the denominator
=√(1−cosA)(1−cosA)(1+cosA)(1−cosA)
=√(1−cosA)21−cos2A
=√(1−cosA)2sin2A
=1−cosAsinA
=1sinA−cosAsinA
=cosec A-cot A=R.H.S
Question 23
(i) √1+sinA1−sinA=tanA+secA
(ii) √1−cosA1+cosA=cosecA−cotA
Sol :
(i) √1+sinA1−sinA=tanA+secA
L.H.S. =√1+sinA1−sinA
Rationalising the denominator
=√(1+sinA)(1+sinA)(1−sinA)(1+sinA)
=√(1+sinA)21−sin2 A
=√(1+sinA)2cos2 A
=1+sinAcosA
=1cosA+sinAcosA
=sec A+ tan A
=tan A+ sec A=R.H.S
(ii) √1−cosA1+cosA=cosecA−cotA
L.H.S. =√1−cosA1+cosA
Rationalising the denominator
=√(1−cosA)(1−cosA)(1+cosA)(1−cosA)
=√(1−cosA)21−cos2A
=√(1−cosA)2sin2A
=1−cosAsinA
=1sinA−cosAsinA
=cosec A-cot A=R.H.S
Question 24
(i) √secA−1secA+1+√secA+1secA−1=2cosecA
(ii) cotAcotA1−sinA=1+cosecA
Sol :
(i) √secA−1secA+1+√secA+1secA−1=2cosecA
LH.S=√secA−1secA+1+√secA+1secA−1
=√secA−1√secA+1+√secA+1√secA−1
=secA−1+secA+1√(secA+1)(secA−1)
=2secA√sec2A−1
{∵sec2A−1=tan2A}
=2secA√tan2A=2secAtanA
=2×cosAcosA×sinA=2sinA
=2cosecA=R.H.S .
(ii) cosAcotA1−sinA=1+cosecA
L.H.S =cosAcotA1−sinA=cosAcosAsinA(1−sinA)
{cosA=cosAsinA}
=cos2AsinA(1−sinA)=1−sin2AsinA(1−sinA)
{∵cos2A=1−sin2A}
=(1+sinA)(1−sinA)sinA(1−sinA)=1+sinAsinA
=1sinA+sinAsinA=cosecA+1
=1+cosec A=R.H.S
Question 25
(i) 1+tanAsinA+1+cotAcosA=2(secA+cosecA)
(ii) sec4A−tan4A=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) sec4A−tan4A=1+2tan2A
L.H.S=sec4A−tan4A
=(sec2A−tan2A)(sec2A+tan2A)
=(1+tan2A−tan2A)(1+tan2A+tan2A)
{∵sec2A=tan2A+1}
=1(1+2tan2A)=1+2tan2A=R.H.S
Question 26
(i) cosec6A−cot6A=3cot2Acosec2A+1
(ii) sec6A−tan6A=1+3tan2A+3tan4A
Sol :
(i) cosec6A−cot6A=3cot2Acosec2A+1
L.H.S=cosec6A−cot6A
=(cosec2A)3−(cot2A)3
=(cosec2θ−cot2A)3+3cosec2Acot2A(cosec2A−cot2A)
=(1)3+3cosec2Acot2A×1
=1+3cot2Acosec2A
=3cot2Acosec2A+1=R.H.S
(ii) sec6A−tan6A=1+3tan2A+3tan4A
L.H.S=sec6A−tan6A
=(sec2A)3−(tan2A)3
=(sec2A−tan2A)3+3sec2Atan2A(sec2A−tan2A)
=(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) cotθ−cosecθ−1cotθ−cosecθ+1=1+cosθsinθ
(ii) sinθcotθ+cosecθ=2+sinθcotθ−cosecθ
Sol :
(i) cotθ−cosecθ−1cotθ−cosecθ+1=1+cosθsinθ
L.H.S =cotθ−cosecθ−1cotθ−cosecθ+1
=cosθsinθ+1sinθ−1cosθsinθ−1sinθ+1
=cosθ+1−sinθsinθ×sinθcosθ−1+sinθ
=cosθ+1−sinθcosθ−1+sinθ
=cosθ+(1−sinθ)cosθ−(1−sinθ)
=[cosθ+(1−sinθ)][cosθ+(1−sinθ)][cosθ−(1−sinθ)][cosθ+(1−sinθ)]
=[cosθ+(1−sinθ)]2cos2θ−(1−sinθ)2
=(cosθ+1−sinθ)2cos2θ−(1+sin2θ−2sinθ)
=cos2θ+sin2θ+1+2cosθ−2sinθ−2sinθcosθcos2θ−1−sin2θ+2sinθ
=1+1+2cosθ−2sinθ−2sinθcosθ1−sin2θ−1−sin2θ+2sinθ
=2+2cosθ−2sinθ−2sinθcosθ2sinθ−2sin2θ
=2(1+cosθ)−2sinθ(1+cosθ)2sinθ(1−sinθ)
=(1+cosθ)2(1−sinθ)2sinθ(1−sinθ)
=1+cosθsinθ=R.H.S
(ii) sinθcotθ+cosecθ=2+sinθcotθ−cosecθ
=sin2θ1+cosθ
=1−cos2θ1+cosθ=(1+cosθ)(1−cosθ)1+cosθ
=1-cos θ
R.H.S=2+sinθcotθ−cosecθ
=2+sinθcosθsinθ−1sinθ
=2+sinθcosθ−1sinθ
=2+sin2θcosθ−1
=2cosθ−2+sin2θcosθ−1
=2cosθ−2+(1−cos2θ)cosθ−1
=2(cosθ−1)+(1+cosθ)(1−cosθ)cosθ−1
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
=(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
(cosecA−sinA)(secA−cosA)sec2A
=(1sinA−sinA)(IcosA−cosA)1cos2 A
=(1−sin2AsinA)(1−cos2AcosA)1cos2A
cos2AsinA⋅sin2AcosA⋅1cos2A=sinAcosA=tanA
R.H.S
(iii) (cosecθ -sinθ)(secθ-cosθ)(tanθ+cotθ)=1
L.H.S=(cosecθ-sinθ)(secθ-cosθ)(tanθ+cotθ)
=(1sinθ−sinθ)(1cosθ−cosθ)(tanθ+cotθ)
=1−sin2θsinθ×1−cos2θcosθ(tanθ+cotθ)
=cos2θsinθ×sin2θcosθ(tanθ+cotθ)
=sinθcosθsin2θ+cos2θsinθcosθ
=1=R.H.S
Question 29
(i) sin3A+cos3AsinA+cosA+sin3A−cos3AsinA−cosA=2
(ii) tan2A1+tan2A+cot2A1+cot2A=1
Sol :
(i) sin3A+cos3AsinA+cosA+sin3A−cos3AsinA−cosA=2
L.H.S=sin3A+cos3AsinA+cosA+sin3A−cos3AsinA−cosA
=(sinA+cosA)(sin2A−sinAcosA+cos2A)(sinA+cosA)+(sinA−cosA)(sin2A+sinAcosA+cos2A)(sinA−cosA)
=(1−sinAcosA)+(1+sinAcosA) [∵sin2A+cos2A=1}
=1−sinAcosA+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+tanA−1cosA=1cosA−1secA−tanA
(ii) (sinA+secA)2+(cosA+cosecA)2=(1+secAcosecA)2
(iii) tanA+sinAtanA−sinA=secA+1secA−1
Sol :
(i) 1secA+tanA−1cosA=1cosA−1secA−tanA
L.H.S=1secA+tanA−1cosA
=11cosA+sinAcosA−1cosA
=cosA1+sinA−1cosA
=cos2A−1−sinAcosA(1+sinA)=−sin2−sinAcosA(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
=\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.
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
\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}
=\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}
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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