Page 4.7 Ex. 4.1
Q1.
Answer :
(i) Let tan-1-3=y
Then,
tany=-3
We know that the range of the principal value branch is -π2, π 2.
Thus,
tany=-3=-tanπ3=tan-π3⇒y=-π3∈-π2,π2
Hence, the principal value of tan-1-3 is -π3.
(ii) Let cos-1-12=y
Then,
cosy=-12
We know that the range of the principal value branch is 0, π.
Thus,
cosy=-12=cos3π4⇒y=3π4∈0,π
Hence, the principal value of cos-1-12 is 3π4.
(iii) Let cosec-1-2=y
Then,
cosecy=-2
We know that the range of the principal value branch is -π2, π2-0.
Thus,
cosecy=-2=cosec-π4y=-π4∈-π2,π2, y≠0
Hence, the principal value of cosec-1-2 is -π4.
(iv) Let cos-1-32=y
Then,
cosy=-32
We know that the range of the principal value branch is 0,π.
Thus,
cosy=-32=cos5π6⇒y=5π6∈0,π
Hence, the principal value of cos-1-32 is 5π6.
(v) Let sin-1-12=y
Then,
siny=-12
We know that the range of the principal value branch is -π2,π2.
Thus,
siny=-12=sin-π6⇒y=-π6∈-π2,π2
Hence, the principal value of sin-1-12 is -π6.
(vi) Let tan-113=y
Then,
tany=13
We know that the range of the principal value branch is -π2,π2.
Thus,
tany=13=tanπ6⇒y=π6∈-π2,π2
Hence, the principal value of tan-113 is π6.
(vii) Let sec-1-2=y
Then,
secy=-2
We know that the range of the principal value branch is 0,π-π2.
Thus,
secy=-2=sec3π4⇒y=3π4∈0,π, y≠π2
Hence, the principal value of sec-1-2 is 3π4.
(viii) Let cot-1-3=y
Then,
coty=-3
We know that the range of the principal value branch is 0,π.
Thus,
coty=-3=cot5π6⇒y=5π6∈0,π
Hence, the principal value of cot-1-3 is 5π6.
(ix) Let sec-12=y
Then,
secy=2
We know that the range of the principal value branch is 0,π-π2.
Thus,
secy=2=secπ3⇒y=π3∈0, π, y≠π2
Hence, the principal value of sec-12 is π3.
(x) Let cosec-123=y
Then,
cosecy=23
We know that the range of the principal value branch is -π2, π2-0.
Thus,
cosecy=23=cosecπ3⇒y=π3∈-π2, π2, y≠0
Hence, the principal value of cosec-123 is π3.
Q2.
Answer :
(i) cos-1cosx=x sin-1sinx=x
cos-112+2sin-112= cos-1cosπ3+2sin-1sinπ6=π3+2π6= 2π3
(ii) Let sin-112=y
Then,
siny=12
We know that the range of the principal value branch is -π2,π2.
Thus,
siny=12=sinπ6⇒y=π6∈-π2,π2
So,
tan-12cos2sin-112=tan-12cos2π6=tan-12cosπ3=tan-1212=tan-11=π4∈-π2,π2
∴tan-12cos2sin-112=π3
(iii) Let sin-1-12=y
Then,
siny=-12
We know that the range of the principal value branch is -π2,π2.
Thus,
siny=-12=sin-π6⇒y=-π6∈-π2,π2
Now,
Let cos-1-12=z
Then,
cosz=-12
We know that the range of the principal value branch is 0,π.
Thus,
cosz=-12=cos2π3⇒z=2π3∈0,π
So,
tan-11+cos-1-12+sin-112=π4+2π3-π6 =3π4
∴tan-11+cos-1-12+sin-112=3π4
(iv) Let tan-13=x
Then,
tanx=3
We know that the range of the principal value branch is -π2,π2.
Thus,
tanx=3=tanπ3⇒x=π3∈-π2,π2
Now,
Let sec-1-2=y
Then,
secy=-2
We know that the range of the principal value branch is 0,π-π2.
Thus,
secy=-2=sec2π3⇒y=2π3∈0,π, y≠π2
Now,
Let cosec-123=z
Then,
cosecz=23
We know that the range of the principal value branch is -π2, π2-0.
Thus,
cosecz=23=cosecπ3z=π3∈-π2, π2, z≠0
So,
tan-13-sec-1-2+cosec-123=π3-2π3+π3=0
∴tan-13-sec-1-2+cosec-123=0
Q3.
Answer :
(i)
sin-112-2sin-112=sin-1sinπ6-2sin-1sinπ4 ∵Range of sine is -π2, π2 and π6, π4∈ -π2, π2=π6-2π4=π6-π2=-π3
∴sin-112-2sin-112=-π3
(ii) sin-1-12+2cos-1-32=sin-1sin-π6+2cos-1cos5π6 ∵Range of sine is -π2, π2 ; -π6∈ -π2, π2 and range of cosine is 0, π ; 5π6∈ 0, π=-π6+25π6=-π6+5π3=9π6=3π2
∴sin-1-12+2cos-1-32=3π2
(iii)
tan-1-1+cos-1-12=tan-1tan-π4+cos-1cos3π4 ∵Range of tan is -π2, π2 ; -π4∈ -π2, π2 and range of cosine is 0, π ; 3π4∈ 0, π=-π4+3π4=π2
∴tan-1-1+cos-1-12=π2
(iv)
sin-1-32+cos-132=sin-1sin-π3+cos-1cosπ6 =-π3+π6 ∵Range of sine is -π2, π2 ; -π3∈ -π2, π2 and range of cosine is 0, π ; π6∈ 0, π=-π6
∴sin-1-32+cos-132=-π6
Page 4.71 Ex. 4.2
Q1.
Answer :
We know
sin θ=θ if -π2≤θ≤π2cos θ=θ if 0≤θ≤πtan θ=θ if -π2<θ<π2
(i) We have
sin-1sin5π6=sin-1sinπ-π6=sin-1sinπ6=π6
(ii) We have
cos-1cos-π4=cos-1cosπ4=π4
(iii) We have
tan-1tan3π4=tan-1tanπ-π4=tan-1-tanπ4=tan-1tan-π4=-π4
(iv) We have
sin-1sin2 =sin-1sinπ-2 ∵ 2∉ -π2,π2 =π-2
(v) We have
sinπ3-sin-1-32=sinπ3-sin-1sin-π3 ∵-π3∈-π2,π2=sinπ3–π3 =sin2π3=32
∴sinπ3-sin-1-32=32
(vi) We have
coscos-1-32+π4=coscos-1cos5π6+π4 =cos5π6+π4 =cos13π12=cosπ+π12=-cosπ12=-cosπ4-π6=-cosπ4cosπ6+sinπ4sinπ6=-12×32+12×12=-3+122
(vii) We have
costan-134=cos12cos-11-3421+342 ∵2tan-1x=cos-11-x21+x2=cos12cos-1725
Let y=cos-1725⇒cosy=725
Now,
cos12cos-1725=cos12y=cosy+12 ∵cos2x=2cos2x-1=725+12=3250=45
∴costan-134=45
(viii) We have
cos-1cos5π4=cos-1cos2π-3π4=cos-1cos3π4=3π4
(ix) We have
cos-1cos4π3=cos-1cos2π-2π3=cos-1cos2π3=2π3
(x) We have
tan-1tan2π3=tan-1tanπ-π3=tan-1-tanπ3=tan-1tan-π3=-π3
(xi) We have
cos-1cos13π6=cos-1cos2π+π6=cos-1cosπ6=π6
(xii) We have
tan-1tan7π6=tan-1tanπ+π6=tan-1tanπ6=π6
Q2.
Answer :
(i)
cossin-135=coscos-11-352 ∵sin-1x=cos-11-x2 =coscos-145 =45
(ii)
sincos-145=sinsin-11-452 ∵cos-1x=sin-11-x2 =sinsin-135 =35
(iii)
cossin-1-35=coscos-11–352 ∵sin-1x=cos-11-x2 =coscos-145 =45
(iv)
tancos-1817=tantan-11-8172817 ∵cos-1x=tan-11-x2x =tantan-11517817 =158
(v)
coseccos-1-1213=coseccosec-111–12132 ∵cos-1x=cosec-111-x2 =coseccosec-11513 =135
(vi)
tan2 tan-115-π4=tan2 tan-115-tan-1 1 =tantan-12×151-152-tan-1 1 ∵2 tan-1x=tan-12×1-x2 =tantan-1252425-tan-1 1 =tantan-1512+tan-1 1 =tantan-1512-11+512 ∵tan-1x-tan-1y=tan-1x-y1+xy =tantan-1-7121712 =tantan-1-717 =-717
(vii)
Let, cos-153=θ⇒cosθ=53⇒2cos2θ2-1=53⇒cos2θ2=3+56⇒cosθ2=3+56⇒θ2=cos-13+56 =tan-11-3+5623+56 =tan-11-3+563+56 =tan-13-563+56 =tan-13-53+5 =tan-13-53-53+53-5 =tan-13-529-5 =tan-13-52i.e., 12cos-153=tan-13-52⇒tan12cos-153=tantan-13-52∴tan12cos-153=3-52
(viii)
sin12cos-145=sin12×2sin-1±1-452 ∵cos-1x=2sin-1±1-x2 =sinsin-1±110 =±110
(ix)
cossin-135+sin-1513=cossin-1351-5132+5131-352 ∵sin-1x+sin-1y=sin-1×1-y2+y1-x2 =cossin-135×1213+513×45 =cossin-13665+413 =cossin-15665 =coscos-11-56652 ∵sin-1x=cos-11-x2 =coscos-13365 =3365
(x)
sin (tan−1 x + cot−1 x)=sinπ2 ∵tan-1x+cot-1x=π2
= 1
Q3.
Answer :
(i)
LHS=cos-1513=tan-11-5132513 ∵cos-1x=tan-11-x2x =tan-11213513 =tan-1125=RHS
(ii)
LHS=sin-1-45=tan-1-451-452 ∵sin-1x=tan-1×1-x2 =tan-1-4535 =tan-1-43=RHSAgain, sin-1-45=-sin-145 =-cos-11-452 ∵sin-1x=cos-11-x2 =-cos-135 =-π-cos-1-35 =cos-1-35-π
(iii)
LHS=tancos-145+tan-123=tantan-11-45245+tan-123 ∵cos-1x=tan-11-x2x =tantan-134+tan-123 =tantan-134+231-34×23 ∵tan-1x+tan-1y=tan-1x+y1-xy =tantan-11712612 =tantan-1176 =176=RHS
(iv)
LHS=2 sin-135=sin-12×351-925 ∵ 2 sin-1x=sin-12×1-x2 =sin-165×45 =sin-12425 =tan-124251-24252 ∵sin-1x=tan-1×1-x2 =tan-12425725 =tan-1247=RHS
(v)
LHS=sin-1513+cos-135 =sin-1513+cos-135 =sin-1513+sin-11-352 ∵sin-1x=cos-11-x2 =sin-1513+sin-145 =sin-15131-452+451-5132 ∵sin-1x+sin-1y=sin-1×1-y2+y1-x2 =sin-1513×35+45×1213 =sin-1313+4865 =sin-16365 =tan-163651-63652 ∵sin-1x=tan-1×1-x2 =tan-163651665 =tan-16316=RHS
Q4.
Answer :
i) LHS=tan-117+tan-1 113 =tan-117+1131-17×113 ∵tan-1x+tan-1y=tan-1x+y1-xy =tan-120919091 =tan-129=RHS
ii) LHS=tan-114+tan-1 29 =tan-114+291-14×29 ∵tan-1x+tan-1y=tan-1x+y1-xy =tan-117363436 =tan-112 =12cos-11-141+14 ∵tan-1x=12cos-11-x21+x2 =12cos-13454 =12cos-135Now,tan-112=12sin-1221+14 ∵tan-1x=12sin-12×1+x2 =12sin-1154 =12sin-145
iii) LHS=tan-123 =12tan-12×231-232 ∵tan-1x=12tan-12×1-x2 =12tan-14359 =12tan-1125=RHS
iv) LHS=tan-117+2tan-1 13 =tan-117+tan-12×131-132 ∵2 tan-1x=tan-12×1-x2 = tan-117+tan-12389 = tan-117+tan-134 =tan-117+341-17×34 ∵tan-1x+tan-1y=tan-1x+y1-xy =tan-125282528 =tan-11=π4=RHS
v) LHS=sin-145+2tan-1 13 =sin-145+tan-12×131-132 ∵2 tan-1x=tan-12×1-x2 = sin-145+tan-12389 = sin-145+tan-134 = sin-145+cos-111+916 ∵tan-1x=cos-111+x2 = sin-145+cos-1154 = sin-145+cos-145 = π2=RHS
vi) LHS=sin-11213+cos-1 45+tan-16316 =tan-112131-144169+tan-11-162545+tan-16316 ∵sin-1x=tan-1×1-x2 and cos-1x=tan-11-x2x =tan-11213513+tan-13545+tan-16316 =tan-1125+tan-134+tan-16316 =π+tan-1125+341-125×34 +tan-16316 ∵tan-1x+tan-1y=π+tan-1x+y1-xy =π+tan-16320-1620+tan-16316 =π+tan-1-6316+tan-16316 =π-tan-16316+tan-16316 =π=RHS
vii) LHS=2 sin-135-tan-1 1731 =2 tan-1341-925-tan-1 1731 ∵sin-1x=tan-1×1-x2 =2 tan-13545-tan-1 1731 =2 tan-134-tan-1 1731 =tan-12×341-342-tan-1 1731 ∵2 tan-1x=tan-12×1-x2 =tan-132716-tan-1 1731 =tan-1247-tan-1 1731 =tan-1247-17311+247×1731 ∵tan-1x-tan-1y=tan-1x-y1+xy =tan-1625217625217 =tan-11=π4=RHS
viii) LHS=cot-17+cot-1 8+cot-118 =tan-117+tan-1 18+tan-1118 ∵cot-1x=tan-11x =tan-117+181-17×18 +tan-1118 ∵tan-1x+tan-1y=tan-1x+y1-xy =tan-115565556+tan-1118 =tan-1311+tan-1118 =tan-1311+1181-311×118 =tan-165198195198 =tan-113 =cot-13=RHS
ix) LHS=2 tan-115+tan-1 18 =tan-12×151-152+tan-1 18 ∵2 tan-1x=tan-12×1-x2 =tan-1252425+tan-1 18 =tan-1512+tan-1 18 =tan-1512+181-512×18 ∵tan-1x+tan-1y=tan-1x+y1-xy =tan-113249196 =tan-147=RHS
x) LHS=2 tan-134-tan-1 1731 =tan-12×341-342-tan-1 1731 ∵2 tan-1x=tan-12×1-x2 =tan-132716-tan-1 1731 =tan-1247-tan-1 1731 =tan-1247-17311+247×1731 ∵tan-1x-tan-1y=tan-1x-y1+xy =tan-1625217625217 =tan-11=π4=RHS
xi) LHS=2 tan-112+tan-1 17 =tan-12×121-122+tan-1 17 ∵2 tan-1x=tan-12×1-x2 =tan-1134+tan-1 17 =tan-143+tan-1 17 =tan-143+171-43×17 ∵tan-1x+tan-1y=tan-1x+y1-xy =tan-131211721 =tan-13117=RHS
Page 4.72 Ex. 4.2
Q5.
Answer :
Let: a=tanm b=tann x=tany
Now,
sin-12a1+a2-cos-11-b21+b2=tan-12×1-x2⇒sin-12tanm1+tan2m-cos-11-tan2n1+tan2n=tan-12tany1-tan2y⇒sin-1sin2m-cos-1cos2n=tan-1tan2y ∵sin2x=2tanx1+tan2x and cos2x=1-tan2x1+tan2x⇒2m-2n=2y⇒m-n=y⇒tan-1a-tan-1b=tan-1x ∵a=tanm, b=tann and x=tany⇒tan-1a-b1+ab=tan-1x ∵tan-1x-tan-1y=tan-1x-y1+xy⇒a-b1+ab=x
∴a-b1+ab=x
Q6.
Answer :
We know
cos-1x+cos-1y=cos-1xy-1-x21-y2
Now,
cos-1×2+cos-1y3=α⇒cos-1x2y3-1-x241-y23=α⇒x2y3-1-x241-y23=cosα⇒xy-4-x29-y2=6cosα⇒4-x29-y2=xy-6cosα⇒4-x29-y2=x2y2+36cos2α-12xycosα Squaring both sides⇒36-4y2-9×2+x2y2=x2y2+36cos2α-12xycosα⇒36-4y2-9×2=36cos2α-12xycosα⇒9×2-12xycosα+4y2=36-36cos2α⇒9×2-12xycosα+4y2=36sin2α
Q7.
Answer :
(i)
tan-11-x22x+cot-11-x22x=π2LHS=tan-11-x22x+cot-11-x22x =tan-11-x22x+π2-tan-11-x22x ∵tan-1x+cot-1x=π2 =π2=RHS
(ii)
sintan-11-x22x+cos-11-x21+x2=1LHS=sintan-11-x22x+cos-11-x21+x2 =sinsin-11-x22x1+1-x22x+cos-11-x21+x2 ∵tan-1x=sin-1×1+x2 =sinsin-11-x21+x2+cos-11-x21+x2 =sinπ2 ∵sin-1x+cos-1x=π2 =1=RHS
(iii)
9π8-94sin-113=94sin-1223LHS=9π8-94sin-113 =94π2-sin-113 =94cos-113 =94sin-11-19 =94sin-1223=RHS
Q8.
Answer :
LHS=2 tan-1a-ba+btanθ2=cos-11-a-ba+btanθ221+a-ba+btanθ22 ∵2 tan-1x=cos-11-x21+x2 =cos-11-a-ba+btan2θ21+a-ba+btan2θ2 =cos-1a+b-a-btan2θ2a+b+a-btan2θ2 =cos-1a+b-atan2θ2+btan2θ2a+b+atan2θ2-btan2θ2 =cos-1a1-tan2θ2+b1+tan2θ2a1+tan2θ2+b1-tan2θ2 =cos-1a1-tan2θ21+tan2θ2+b1+tan2θ21+tan2θ2a1+tan2θ21+tan2θ2+b1-tan2θ21-tan2θ2 Dividing Nr and Dr by 1+tan2θ2 =cos-1a1-tan2θ21+tan2θ2+ba+b1-tan2θ21-tan2θ2 =cos-1acosθ+ba+bcosθ=RHS
Q9.
Answer :
Let:
a=tanzb=tany
Then,
sin-12a1+a2+sin-12b1+b2=2tan-1x⇒sin-12tanz1+tan2z+sin-12tany1+tan2y=2tan-1x⇒sin-1sin2z+sin-1sin2y=2tan-1x ∵sin2x=2tanx1+tan2x⇒2z+2y=2tan-1x⇒tan-1a+tan-1b=tan-1x ∵a=tanz and b=tany⇒tan-1a+b1-ab=tan-1x ∵tan-1x+tan-1y=tan-1x+y1-xy ⇒x=a+b1-ab
Q10.
Answer :
We have
2tan-1x+sin-12×1+x21 For x>1,=2tan-1x+sin-12×1+x2=π-sin-12×1+x2+sin-12×1+x2 ∵ 2tan-1x=π-sin-12×1+x2 , x>1=π2 For x=1,=2tan-1x+sin-12×1+x2 =2tan-11+sin-1211+12=2tan-11+sin-11=2π4+π2 =π
Q11.
Answer :
We know
tan-1x+tan-1y=tan-1x+y1-xy, xy>1
∴ tan-12aba2-b2+tan-12xyx2-y2=tan-12aba2-b2+2xyx2-y21-2aba2-b22xyx2-y2=tan-12abx2-aby2+xya2-xyb2a2-b2x2-y2a2x2-a2y2-x2b2+y2b2-4abxya2-b2x2-y2=tan-12abx2-aby2+xya2-xyb2a2x2-a2y2-x2b2+y2b2-2abxy-2abxy=tan-12ax-byay+bxax-by2-ay+bx2=tan-12αβα2-β2 ∵ α=ax-by and β=ay+bx
Q12.
Answer :
(i)
sin-1×1-y2-y1-x2=sin-1x-sin-1yPutting y=x, sin-1×1-x2-x1-x2=sin-1x-sin-1x⇒sin-1×1-x-x1-x2=sin-1x-sin-1x
(ii) Let x=cotθ
Now,
tan-1x+1+x2=tan-1cotθ+1+cot2θ=tan-1cotθ+cosecθ =tan-1cosθ+1sinθ=tan-12cos2θ22sinθ2cosθ2=tan-1cotθ2=tan-1tanπ2-θ2 =π2-θ2=π2-cot-1×2
(iii) Let x=cotθ
Now,
tan-11+x2-x=tan-11+cot2θ-cotθ=tan-1cosecθ-cotθ=tan-11-cosθsinθ=tan-12sin2θ22sinθ2cosθ2=tan-1tanθ2=θ2=cot-1×2
(iv) Let x=tanθ
Now,
tan-11+x2-1x=tan-11+tan2θ-1tanθ=tan-1sec2θ-1tanθ =tan-1secθ-1tanθ=tan-11-cosθsinθ=tan-12sin2θ22sinθ2cosθ2=tan-1tanθ2=θ2=tan-1×2
(v) Let x=tanθ
Now,
tan-11+x2+1x=tan-11+tan2θ+1tanθ=tan-1sec2θ+1tanθ =tan-1secθ+1tanθ=tan-1cosθ+1sinθ=tan-12cos2θ22sinθ2cosθ2=tan-1cotθ2=tan-1tanπ2-θ2 =π2-θ2=π2-tan-1×2
(vi) Let x=acosθ
Now,
tan-1a-xa+x=tan-1a-acosθa+acosθ=tan-11-cosθ1+cosθ=tan-12sin2θ22cos2θ2=tan-1tanθ2=θ2 =12cos-1xa
∴ tan-1a-xa+x=cos-1xa2
(vii) Let x=asinθ
Now,
tan-1xa+a2-x2=tan-1asinθa+a2-a2cos2θ=tan-1asinθa+acos2θ=tan-1sinθ1+cosθ=tan-12sinθ2cosθ22cos2θ2=tan-1tanθ2 =θ2=12sin-1xa
(viii) Let x=sinθ
Now,
sin-1x+1-x22=sin-1sinθ+1-sin2θ2=sin-1sinθ+cosθ2=sin-112sinθ+12cosθ=sin-1cosπ4sinθ+sinπ4cosθ=sin-1sinθ+π4 =θ+π4=π4+sin-1x
∴ sin-1x+1-x22=cos-1x+π4
(ix) Let x=cosθ
Now,
sin-11+x+1-x2=sin-11+cosθ+1-cosθ2=sin-12cos2θ2+2sin2θ22=sin-1cosθ2+sinθ22=sin-112sinθ2+12cosθ2=sin-1sinθ2+π4 =θ2+π4=cos-1×2+π4
∴ sin-11+x+1-x2=cos-1×2+π4
(x) Let x=cosθ
Now,
sin2tan-11-x1+x=sin2tan-11-cosθ1+cosθ=sin2tan-12sin2θ22cos2θ2=sin2tan-1tanθ2=sinθ=sincos-1x =sinsin-11-x2=1-x2
(xi) Let x=asecθ
Now,
cot-1ax2-a2=cot-1aa2sec2θ-a2=cot-1aatan2θ=cot-1cotθ=θ =sec-1xa
Page 4.73 Ex. 4.2
Q13.
Answer :
(i) Let sin-112=y
Then,
siny=12
∴ tan-12cos2sin-112=tan-12cos2y=tan-121-2sin2y ∵cos2x=1-2sin2x=tan-121-2×14 ∵siny=12=tan-12×12=tan-11=π4
∴ tan-12cos2sin-112=π4
(ii)
We havecottan-1a+cot-1a=cotπ2 tan-1x+cot-1x=π2=0
∴ cottan-1+cot-1a=0
(iii)
We have
cossec-1x+cosec-1x=cosπ2 ∵sec-1x+cosec-1x=π2=0
∴ cossec-1x+cosec-1x=0 , |x|≥1
Q14.
Answer :
Let a=btanm and x=ytann
Then,
23tan-13ab2-a3b3-3a2b+23tan-13xy2-x3y3-3x2y=23tan-13b3tanm-b3tan3mb3-3b3tan2m+23tan-13y3tann-y3tan3ny3-3y3tan2n=23tan-13tanm-tan3m1-3tan2m+23tan-13tann-tan3n1-tan2n=23tan-1tan3m+23tan-1tan3n ∵ tan3x=3tanx-tan3x1-3tan2x=233m+233n =2m+2n=2tan-1ab+tan-1xy ∵ a=btanm, x=ytann=2tan-1ab+xy1-abxy=2tan-1ay+bxby-ax=tan-12ay+bxby-ax1-ay+bxby-ax2=tan-12ay+bxby-axby-ax2-ay+bx2=tan-12αβα2-β2 ∵ β=ay+bx and α=by-ax
Q15.
Answer :
(i) We know
tan-1x+tan-1y=tan-1x+y1-xy
∴ tan-12x+tan-13x=nπ+3π4⇒tan-12x+3×1-2x×3x=nπ+3π4⇒5×1-6×2=tannπ+3π4⇒5×1-6×2=-1⇒5x=-1+6×2⇒6×2-5x-1=0⇒6x+1x-1=0⇒x=-16 As x=1 is not satisfying the equation
(ii) We know
tan-1x+tan-1y=tan-1x+y1-xy
∴ tan-1x+1+tan-1x-1=tan-1831⇒tan-1x+1+x-11-x+1×x-1=tan-1831⇒2×1-x2+1=831⇒2×2-x2=831⇒31x=8-4×2⇒4×2+31x-8=0⇒4×2+32x-x-8=0⇒4x-1x+8=0⇒x=14 As x=-8 is not satisfying the equation
(iii) We know
tan-1x+tan-1y=tan-1x+y1-xy
∴ tan-114+2tan-115+tan-116+tan-11x=π4⇒tan-114+tan-115+tan-115+tan-116+tan-11x=π4⇒tan-114+151-14×15+tan-115+161-15×16+tan-11x=π4⇒tan-19201920+tan-111302930+tan-11x=π4⇒tan-1919+tan-11129+tan-11x=π4⇒tan-1919+11291-1129×919+tan-11x=π4⇒tan-1235226+tan-11x=π4⇒tan-1235226+1×1-235226×1x=π4⇒235x+226226x-235=tanπ4⇒235x+226226x-235=1⇒235x+226=226x-235⇒9x=-461⇒x=-4619
(iv) We know
sin-1x+sin-1y=sin-1×1-y2+y1-x2
∴ sin-1x+sin-12x=π3⇒sin-1x+sin-12x=sin-132⇒sin-1x-sin-132=-sin-12x⇒sin-1×1-34+321-x2=-sin-12x⇒sin-1×2+321-x2=sin-1-2x⇒x2+321-x2=-2x⇒x+31-x2=-4x⇒5x=-31-x2Squaring both the sides,25×2=3-3×2⇒28×2=3⇒x=±1237
(v)
3sin-12×1+x2-4cos-11-x21+x2+2tan-12×1-x2=π3⇒6tan-1x-8tan-1x+4tan-1x=π3 ∵ 2tan-1x=sin-12×1+x2, 2tan-1x=cos-11-x21+x2 and 2tan-1x=tan-12×1-x2 ⇒2tan-1x=π3⇒tan-1x=π6 ⇒x=tanπ6⇒x=13
∴ x=13
(vi)
cos-1x+sin-1×2=π6⇒cos-1x+sin-1×2=sin-112⇒cos-1x=sin-112-sin-1×2⇒cos-1x=sin-1121-x24-x21-14 ∵ sin-1x-sin-1y=sin-1×1-y-y1-x2⇒cos-1x=sin-13×4-3×4⇒sin-11-x2=sin-13×4-3×4⇒1-x2=0Squaring both the sides,⇒1-x2=0 ⇒x=±1 As x=-1 is not satisfying the equation
(vii) We know
tan-1x+tan-1y=tan-1x+y1-xy and tan-1x-tan-1y=tan-1x-y1+xy
∴ tan-1x+1+tan-1x-1+tan-1x=tan-13x⇒tan-1x+1+x-11-x+1×x+1=tan-13x-tan-1x⇒tan-12×2-x2=tan-13x-x1+3×2⇒2×2-x2=2×1+3×2⇒2-x2=1+3×2⇒4×2-1=0⇒x2=14⇒x=±12
(viii)
tancos-1x=sincot-112 ⇒tancos-1x=sintan-12 ∵ cot-1x=tan-11x⇒tantan-11-x2x=sinsin-121+4 ∵ cos-1x=tan-11-x2x and tan-1x=sin-1×1+x2⇒1-x2x=25Squaring both the sides,⇒1-x2x2=45⇒5-5×2=4×2⇒9×2=5⇒x=±53
(ix)
tan-11-x1+x-12tan-1x=0⇒tan-11-x1+x=12tan-1x⇒tan-11- tan-1x=12tan-1x ∵ tan-11- tan-1x=tan-11-x1+x⇒tan-11=32tan-1x⇒π4=32tan-1x⇒π6=tan-1x⇒x=13
(x)
⇒cot-1x-cot-1x+2=π12⇒tan-11x+cot-11x+2=π12 ∵ cot-1x=tan-11x⇒tan-11x-1x+21+1xx+2=π12 ⇒tan-12xx+2×2+2x+1xx+2=π12⇒tan-12×2+2x+1=π12⇒2×2+2x+1=tanπ12 ⇒2×2+2x+1=tanπ3-π4 ⇒2×2+2x+1=tanπ3-tanπ41+tanπ3×tanπ4⇒2×2+2x+1=3-13+1⇒2×2+2x+1=3-13+1×3+13+1⇒2×2+2x+1=23+12⇒1x+12=13+12⇒x+1=3+1⇒x=3
(xi) We know
tan-1x+tan-1y=tan-1x+y1-xy
∴ tan-12×1-x2+cot-11-x22x=2π3⇒tan-12×1-x2+tan-12×1-x2=2π3 ∵ cot-1x=tan-11x⇒tan-12×1-x2=π3⇒2tan-1x=π3 ∵ 2tan-1x tan-12×1-x2⇒tan-1x=π6⇒x=tanπ6⇒x=13
(xii) We know
tan-1x+tan-1y=tan-1x+y1-xy
∴ tan-1x+2+tan-1x-2=tan-1879⇒tan-1x+2+x-21-x+2×x-2=tan-1879⇒2×1-x2+4=879⇒x5-x2=479⇒79x=20-4×2⇒4×2+79x-20=0⇒4×2+80x-x-20=0⇒4x-1x+20=0⇒x=14 or- 20∴ x=14 ∵ x>0
(xiii) We know
tan-1x+tan-1y=tan-1x+y1-xy
∴ tan-1×2+tan-1×3=π4⇒tan-1×2+x31-x2×x3=π4⇒tan-15×66-x26=π4⇒5×6-x2=tanπ4⇒5×6-x2=1⇒5x=6-x2⇒x2+5x-6=0⇒x-1x+6=0⇒x=1 ∵ 0<x<6
Q16.
Answer :
We know
tan-1x-tan-1y=tan-1x-y1+xy,xy>-1
Now,
tan-1xy-tan-1x-yx+y
=tan-1xy-x-yx+y1+xyx-yx+y=tan-1×2+xy-xy+y2yx+yx2+y2+xy-xyyx+y=tan-11=tan-1tanπ4=π4
∴ tan-1xy-tan-1x-yx+y=π4
Page 4.76 (Very Short Answers)
Q1.
Answer :
sin-1-x=-sin-1x, x∈-1,1cos-1-x=π-cos-1x, x∈-1,1
∴ sin-1-32+cos-1-12=-sin-132+π-cos-112 = -sin-1sinπ3+π-cos-1cosπ3=-π3+π-π3=π3
∴ sin-1-32+cos-1-12=π3
Q2.
Answer :
The maximum value of sin-1x in x∈-1,1 is at 1.
So, the maximum value is
sin-11 = sin-1sinπ2=π2
Again, the minimum value is at -1.
Thus, the minimum value is
sin-1-1=-sin-11 = -sin-1π2=-π2
So, the difference between the maximum and the minimum value is
π2–π2=π
Q3.
Answer :
sin-1x+sin-1y+sin-1z=3π2⇒sin-1x+sin-1y+sin-1z=π2+π2+π2 As the maximum value in the range of sin-1x is π2And here sum of three inverse of sine is 3 times π2. i.e., every sin inverse function is equal to π2 here.⇒sin-1x=π2, sin-1y=π2 and sin-1z=π2⇒x=1, y=1 and z=1∴ x+y+z=1+1+1=3
Q4.
Answer :
sin-12×1+x2=π-2tan-1x ∵ 2tan-1x=π- sin-12×1+x2 for x>1
Q5.
Answer :
Let x=tany
Then,
cos-11-x21+x2=cos-11-tan2y1+tan2y=cos-1cos2y ∵ 1-tan2x1+tan2x=cos2x=2y …1
The value of x is negative.
So, let x = -a where a > 0.
-a=tany⇒y=tan-1-a
Now,
cos-11-x21+x2=2y Using 1=2tan-1-a =-2tan-1x ∵ x=-a
Q6.
Answer :
tan-1x+tan-1y=tan-1x+y1-xy, xy<1
∴ tan-1x+tan-11x=tan-1x+1×1-x1x,x>0=tan-1×2+10=tan-1∞=tan-1tanπ2=π2
∴ tan-1x+tan-11x=π2
Q7.
Answer :
tan-1x+tan-1y=tan-1x+y1-xy
When x<0, 1x<0, then both are negative.
Let x = – y, y>0
Then,
tan-1x+tan-11x=tan-1-y+tan-1-1y=-tan-1y+tan-11y=-tan-1y+1y1-y1y,y>0=-tan-1y2+10=-tan-1∞=-tan-1tanπ2=-π2
∴ tan-1x+tan-11x=-π2, x<0
Q8.
Answer :
cos-1cos 2π3+sin-1sin 2π3=cos-1cos 2π3+sin-1sin π-π3 =cos-1cos 2π3+sin-1sin π3 ∵ Range of sine is -π2, π2 ; π3∈ -π2, π2 and range of cosine is 0, π ; 2π3∈ 0, π =2π3+π3=π
Page 4.77 (Very Short Answers)
Q9.
Answer :
Let x=-tany
where 0<y<π2
Then,
sin-12×1+x2+cos-11-x21+x2=sin-1-2tany1+tan2y+cos-11-tan2y1+tan2y=sin-1-sin 2y+cos-1cos2y =-sin-1sin 2y+cos-1cos2y =-2y+2y =0
∴ sin-12×1+x2+cos-11-x21+x2=0
Q10.
Answer :
We know
cot-1x=tan-11x
Now, we have
sincot-1x=sintan-11x=sinsin-11×1+1×2 ∵ tan-1x=sin-1×1+x2=sinsin-11xx2+1x=sinsin-1 1×2+1=1×2+1 ∵ sinsin-1x=x
Hence, sincot-1x=1×2-1
Q11.
Answer :
We have
cos-112+2sin-112=cos-1cosπ3+2sin-1sinπ6 ∵The range of sine is -π2, π2; π6∈ -π2, π2 and the range of cosine is 0, π; π3∈ 0, π =π3+2π6=π3+π3=2π3
∴ cos-112+2sin-112=2π3
Q12.
Answer :
The range of tan-1x is-π2,π2.
Q13.
Answer :
We know that
cos-1cosx=x
Now,
cos-1cos1540∘=cos-1cos1440+100∘=cos-1cos100∘ ∵cos4π+100∘=cos100°=100∘
Q14.
Answer :
We know that sin-1sinx=x.
Now,
sin-1sin-600∘=sin-1sin720∘-600∘=sin-1sin120∘=sin-1sin180∘-120∘ ∵sinx = sinπ-x=sin-1sin60∘=60∘
∴ sin-1sin-600∘=60∘
Q15.
Answer :
Let y=sin-113
Then, siny=13
Now, cosy=1-sin2y,
⇒cosy=1-19=89=223
cos2sin-113=cos(2y)=cos2y-sin2y ∵cos 2x= cos2x-sin2x=2232-132=89-19=79
∴ cos2sin-113=79
Q16.
Answer :
We know that sin-1sinx=x.
Now,
sin-1sin1550∘=sin-1sin1620∘-1550∘ ∵sinx =sin1620∘-x=sin-1sin70∘=70∘
∴ sin-1sin1550∘=70∘
Q17.
Answer :
We know that
cos-1x=2tan-11-x1+xtan-1x=sin-1×1+x2
∴ sin12cos-145=sin122tan-11-451+45=sintan-11595=sintan-113=sinsin-1131+19=sinsin-1110=110 ∵sinsin-1x=x
∴ sin12cos-145=110
Q18.
Answer :
We know that
tan-1x=sin-1×1+x2
∴ sintan-134=sinsin-1341+916=sinsin-13454=sinsin-135=35 ∵sinsin-1x=x
∴ sintan-134=35
Q19.
Answer :
We have
cos-1tan3π4=cos-1-tanπ-3π4 ∵tanπ-x=-tanx=cos-1tan-π4=cos-1-tanπ4=cos-1-1=cos-1cosπ ∵cosπ=-1 =π
∴ cos-1tan3π4=π
Q20.
Answer :
We have, cos 2 sin-112 = cos2×π6=cosπ3=12
Q21.
Answer :
We have
cos-1cos350∘-sin-1sin350∘=cos-1cos360∘-350∘-sin-1sin360∘-350∘ ∵sin360∘-x=-sinx , cos360∘-x=cosx =cos-1cos10∘-sin-1sin-10∘=10∘–10∘=20∘
∴ cos-1cos350∘-sin-1sin350∘=20∘
Q22.
Answer :
Let y=cos-135⇒cosy=35
Now,
cos212cos-135=cos212y=cosy+12 ∵cos2x = 2cos2x-1=35+12=852=45
∴ cos212cos-135=45
Q23.
Answer :
We know that tan-1x+tan-1y=tan-1x+y1-xy.
Now,
tan-1x+tan-1y=π4⇒tan-1x+y1-xy=π4⇒x+y1-xy=tanπ4⇒x+y1-xy=1 ⇒x+y=1-xy⇒x+y+xy=1
∴ x+y+xy=1
Q24.
Answer :
We know that cos-1cosx=x.
Now,
cos-1cos6=cos-1cos2π-6=2π-6
Q25.
Answer :
Consider,
sin-1cosπ9=sin-1sinπ2-π9 ∵cos x =sinπ2-x=sin-1sin7π18=7π18 ∵ sin-1sinx =x
∴ sin-1cosπ9=7π18
Q26.
Answer :
We have
sinπ3-sin-1-12=sinπ3–π6=sinπ3+π6=sinπ2=1
∴ sinπ3-sin-1-12 = 1
Q27.
Answer :
We have
tan-1tan15π4=tan-1tan4π-π4tan-1-tanπ4 ∵tan4π-x=-tanx=tan-1tan-π4 =-π4 ∵tan-1tanx =x
∴ tan-1tan15π4=-π4
Q28.
Answer :
cos-1-x=π-cos-1 x, x∈-1,1
Now, we have
sin-112+cos-1-12=sin-112+π-cos-112=sin-1sinπ6+π-cos-1cosπ3=π6+π-π3=5π6
∴ sin-112+cos-1-12=5π6
Q29.
Answer :
We know that tan-1x-tan-1y=tan-1x-y1+xy.
Now,
tan-1ab-tan-1a-ba+b=tan-1ab-a-ba+b1+aba-ba+b=tan-1a2+ab-ab+b2ba+bab+b2-ab+a2ba+b=tan-11=tan-1tanπ4 ∵tanπ4=1=π4
∴ tan-1ab-tan-1a-ba+b=π4
Q30.
Answer :
cos-1cos5π4≠5π4 as 5π4 does not lie between 0 and π.
We have
cos-1cos5π4=cos-1cos2π-3π4=cos-1cos3π4=3π4
Q31.
Answer :
We have
LHS=sin-12×1-x2Putting x=sin a, we get=sin-12 sin a1-sin2a =sin-12sin a cos a=sin-1sin 2a=2a=2sin-1x ∵x=sin a
Q32.
Answer :
We know that sin-1sinx=x.
We have
sin-1sin3π5=sin-1sinπ-3π5 ∵π-3π5∈-π2,π2=sin-1sin2π5=2π5
∴ sin-1sin3π5=2π5
Q33.
Answer :
We know that tan-1x+cot-1x=π2.
We have
tan-13+cot-1x=π2⇒tan-13=π2-cot-1x⇒tan-13=tan-1x⇒x=3
∴x=3
Q34.
Answer :
We know that sin-1x+cos-1x=π2.
We have
sin-113+cos-1x=π2⇒sin-113=π2-cos-1x⇒sin-113=sin-1x⇒x=13
∴ x=13
Page 4.78 (Very Short Answers)
Q35.
Answer :
We know that sin-1x+cos-1x=π2 and cos-1-x=π-cos-1x.
∴ sin-113-cos-1-13=sin-113-π-cos-113=sin-113-π+cos-113=sin-113+cos-113-π=π2-π ∵sin-1x+cos-1x=π2=-π2
∴ sin-113-cos-1-13=-π2
Q36.
Answer :
We know that sin-1x+cos-1x=π2
∴ 4sin-1x+cos-1x=π⇒4sin-1x+π2-sin-1x=π ∵sin-1x+cos-1x=π2⇒3sin-1x=π 2⇒sin-1x=π 6⇒x=sinπ 6⇒x=12
∴ x=12
Q37.
Answer :
We know
tan-1x+tan-1y=tan-1x+y1-xy
x<0, y<0 such that
xy=1
Let x = -a and y = -b where both a and b are positive.
∴ tan-1x+tan-1y=tan-1x+y1-xy=tan-1-a-a1-1=tan-1-∞=tan-1tan-π2=-π2
Q38.
Answer :
Let y=sin-1-32
Then,
siny=-32=sin-π3y=-π3∈-π2,π2
Here, -π2,π2 is the range of the principal value branch of inverse sine function.
∴ sin-1-32=-π3
Q39.
Answer :
Let y=sin-1-12
Then,
siny=-12=sin-π6y=-π6∈-π2,π2
Here, -π2,π2 is the range of the principal value branch of the inverse sine function.
∴ sin-1-12=-π6
Q40.
Answer :
We have, cos-1cos2π3+sin-1sin2π3=cos-1cos2π3+sin-1sinπ-π3 ∵ π-2π3∈-π2,π2=cos-1cos2π3+sin-1sinπ3=2π3+π3=π
∴ cos-1cos2π3+sin-1sin2π3=π
Page 4.78 (Multiple Choice Questions)
Q1.
Answer :
(a) sin 2α
tan-11+x2-1-x21+x2+1-x2=α⇒1+x2-1-x21+x2+1-x2=tanα⇒ 1+x2-1-x21+x2+1-x2×1+x2-1-x21+x2-1-x2 =tanα⇒1+x22+1-x22-21+x21-x21+x22-1-x22=tanα⇒1-1-x4x2=tanα⇒x2tanα=1-1-x4⇒1-x4=1-x2tanα⇒1-x4=1+x4tan2α-2x2tanα⇒x4+x4tan2α-2x2tanα=0⇒x4sec2α-2x2tanα=0⇒x2x2sec2α-2tanα=0⇒x2sec2α-2tanα=0 ∵x2≠0⇒x2sec2α=2tanα⇒x2=2tanαsec2α=2sinαcosα=sin2α
Q2.
Answer :
(d) 329
Let, cos-1152=y and sin-1417=z
∴ cosy=152⇒siny=752⇒tany=7sinz=417⇒cosz=117⇒tanz=4
∴ tancos-1152-sin-1417=tany-z=tany-tanz1+tany tanz=7-41+7×4=329
Q3.
Answer :
(c) tan-1 x
Let tan-1x=y
So, x=tany
∴ 2tan-1cosectan-1x-tancot-1x=2tan-1cosectan-1x-tantan-11x =2tan-1cosectan-1x-1x=2tan-1cosec y-1tany=2tan-11-cosysiny=2tan-12sin2y2siny =2tan-12sin2y22siny2cosy2=2tan-1tany2=y=tan-1x
Q4.
Answer :
(a) sin2 α
We know that cos-1x+cos-1y=cos-1xy-1-x21-y2.
∴ cos-1xa+cos-1yb=α⇒cos-1xayb-1-x2a21-y2b2=α⇒xyab-1-x2a21-y2b2=cosα⇒1-x2a21-y2b2=xyab-cosα⇒1-x2a21-y2b2=x2a2y2b2+cos2α-2xyabcosα Squaring both the sides⇒1-x2a2-y2b2+x2a2y2b2=x2a2y2b2+cos2α-2xyabcosα⇒x2a2+y2b2-2xyabcosα=1-cos2α=sin2α
Q5.
Answer :
(a) x = 1, y = 2
We have,tan-1x+cos-1y1+y2=sin-1310⇒tan-1x+tan-11-y1+y22y1+y2=tan-13101-3102⇒tan-1x+tan-11y=tan-13⇒tan-1x+1y1-x×1y=tan-13⇒xy+1y-x=3⇒3y-3x=xy+1⇒3x+xy=3y-1⇒x3+y=3y-1⇒x=3y-13+yFor, y=1 ⇒x=12For, y=2 ⇒x=1For, y=3 ⇒x=43For, y=4 ⇒x=117For, y=1 ⇒x=73 and so on……Therefore, only integral solutions are :x=1 and y=2
Page 4.79 (Multiple Choice Questions)
Q6.
Answer :
(b) 32
We know that sin-1x+cos-1x=π2.
∴ sin-1x-cos-1x=π6⇒π2-cos-1x -cos-1x=π6⇒-2cos-1x =π6-π2⇒-2cos-1x =-π3⇒cos-1x =π6⇒x=cosπ6⇒x=32
Q7.
Answer :
(a) x
Let cos-1x=y
Then,
sincot-1tancos-1x=sincot-1tan y =sincot-1cot π2-y =sinπ2-y=cosy =x ∵cosy=x
Q8.
Answer :
(a) 2
We know that tan-1x+tan-1y=tan-1x+y1-xy.
∴ tan-12x+tan-13x=π4⇒tan-12x+3×1-2x ×3x=π4⇒2x+3×1-2x ×3x=tanπ4⇒5×1-6×2=1 ⇒5x=1-6×2⇒6×2+5x-1=0
Therefore, there are two solutions.
Q9.
Answer :
(a) 4 α = 3 β
We know that tan-1tanx=x.
∴ α=tan-1tan5π4=tan-1tanπ+π4=tan-1tanπ4=π4
and
β=tan-1-tan2π3=tan-1-tanπ-π3=tan-1tanπ3=π3
∴ 4α=π3β=π
∴ 4α = 3β
Q10.
Answer :
(c) 2
For, -π≤x≤-π21+cos 2x=2sin-1(sin x)⇒2 cos x=2 -π-x⇒2 -cos x=2 -π-x⇒cosx=π+x It does not satisfy for any value of x in the interval -π, -π2
For, -π2≤x≤π21+cos 2x=2sin-1(sin x)⇒2 cos x=2 x⇒2 cos x=2 x⇒cosx= x It gives one value of x in the interval -π2, π2
For, π2≤x≤π1+cos 2x=2sin-1(sin x)⇒2 cos x=2 -π-x⇒2 -cos x=2 π-x⇒cosx=-π+x It gives one value of x in the interval π2, π
∴1+cos 2x=2sin-1(sin x) gives two real solutions in the interval -π, π
Q11.
Answer :
(b) -π2
We know that tan-1x+tan-1y=tan-1x+y1-xy.
x<0, y<0 such that
xy=1
Let x = -a and y = -b, where a and b both are positive.
∴ tan-1x+tan-1y=tan-1x+y1-xy=tan-1-a-a1-1=tan-1-∞=tan-1tan-π2=-π2
Q12.
Answer :
(a) tan θ
Let y=tanθ
Then,
u=cot-1tanθ-tan-1tanθ⇒u=cot-1y-tan-1y⇒u=π2-2tan-1y ∵ tan-1x+cot-1x=π2 ⇒2tan-1y=π2-u ⇒tan-1y=π4-u2⇒y=tanπ4-u2⇒tanθ=tanπ4-u2 ∵y=tanθ
Q13.
Answer :
(c) 18 − 18 cosθ
We know
cos-1x+cos-1y=cos-1xy-1-x21-y2
∴ cos-1×3+cos-1y2=θ2⇒cos-1x3y2-1-x291-y24=θ2⇒xy6-9-x294-y24=cosθ2⇒xy-6cosθ2=9-x2 4-y2
Squaring both the sides, we get
x2y2-12xycosθ2+36cos2θ2=9-x24-y2⇒x2y2-12xycosθ2+36cos2θ2=36-9y2-4×2+x2y2⇒4×2+9y2-12xycos2θ2=36-36cos2θ2⇒4×2+9y2-12xycos2θ2=361-cosθ+12 ∵cos2x =2cos2x-1⇒4×2+9y2-12xycos2θ2=18-18cosθ
Q14.
Answer :
(a) π6
We have
α = tan-13x2y-x, β=tan-12x-y3y
Now, α-β=tan-13x2y-x-tan-12x-y3y =tan-13x2y-x-2x-y3y1+3x2y-x×2x-y3y =tan-13xy-4xy+2y2+2×2-xy3y2y-x3y2y-x+3x2x-y3y2y-x =tan-13xy-4xy+2y2+2×2-xy23y2-3xy+23×2-3xy =tan-12y2+2×2-2xy23y2+23×2-23xy =tan-113=π6
Q15.
Answer :
(b) e13π/18
Given: fx=ecos-1sinx+π3
Then,
f8π9=ecos-1sin8π9+π3=ecos-1sin11π9=ecos-1cosπ2+13π18 ∵cosπ2+θ=sinθ=ecos-1cos13π18=e13π18
Q16.
Answer :
(d) none of these
We know that tan-1x+tan-1y=tan-1x+y1-xy.
Now,
tan-1111+tan-1211=tan-1111+2111-111211=tan-1311121-2121=tan-1311119121=tan-133119=0.27
Q17.
Answer :
(c) 36 sin2 θ
We know
cos-1x+cos-1y=cos-1xy-1-x21-y2
Now,
cos-1×2+cos-1y3=θ⇒cos-1x2y3-1-x241-y23=θ⇒x2y3-1-x241-y23=cosθ⇒xy-4-x29-y2=6cosθ⇒4-x29-y2=xy-6cosθ⇒4-x29-y2=x2y2+36cos2θ-12xycosθ (Squaring both the sides)⇒36-4y2-9×2+x2y2=x2y2+36cos2θ-12xycosθ⇒36-4y2-9×2=36cos2θ-12xycosθ⇒9×2-12xycosθ+4y2=36-36cos2θ⇒9×2-12xycosθ+4y2=36sin2θ
Q18.
Answer :
(b) 15
We know that tan-1x+tan-1y=tan-1x+y1-xy.
Now,
tan-13+tan-1x=tan-18⇒tan-13+x1-3x=tan-18⇒3+x1-3x=8⇒3+x=8-24x⇒3-8=-24x-x⇒-5=-25x⇒x=525=15
Q19.
Answer :
(b) -π10
sin-1cos33π5=sin-1cos6π+3π5=sin-1cos3π5=sin-1sinπ2-3π5=π2-3π5=-π10
Page 4.80 (Multiple Choice Questions)
Q20.
Answer :
(d) 0
We have
cos-1cos5π3+sin-1sin5π3=cos-1cos2π-π3+sin-1sin2π-π3=cos-1cosπ3+sin-1-sinπ3=cos-1cosπ3-sin-1sinπ3=π3-π3=0
Q21.
Answer :
(d) -2425
Let cos-1-35=x, 0≤x≤π
Then, cos x=-35
∴ sinx=1-cos2x=1–352=1625=45
Now,
sin2cos-1-35=sin2x=2sinx cosx=2×45×-35=-2425
Q22.
Answer :
(a) π3
We know
sin-1sinx=x
Now,
θ=sin-1sin-600∘=sin-1sin720∘-600∘=sin-1sin120∘=sin-1sin180∘-120∘ ∵sinx = sinπ-x=sin-1sin60∘=60∘
Q23.
Answer :
Answer :
(a) 13
Let x=tany
Then,
3sin-12tany1+tan2y-41-tan2y1+tan2y+2tan-12tany1-tan2y=π3⇒3sin-1sin 2y-4cos-1cos 2y+2tan-1tan2y=π3 ∵sin2y=2tany1+tan2y,cos2y=1-tan2y1+tan2y and tan2y=2tany1-tan2y⇒3×2y-4×2y+2×2y=π3⇒6y-8y+4y=π3⇒2y=π3⇒y=π6⇒tan-1x=π6 ∵tan-1x=y⇒x=tanπ6⇒x=13
Q24.
Answer :
(c) 32
We know that sin-1x+cos-1x=π2.
4cos-1x+sin-1x=π⇒4cos-1x+π2-cos-1x=π⇒3cos-1x=π-π2⇒3cos-1x=π2⇒cos-1x=π6⇒x=cosπ6⇒x=32
Q25.
Answer :
(d) 2
We know that tan-1x+tan-1y=tan-1x+y1-xy.
∴ tan-1x+1x-1+tan-1x-1x=tan-1-7⇒tan-1x+1x-1+x-1×1-x+1x-1×x-1x=tan-1-7⇒tan-1×2+x+x2-2x+1xx-1×2-x-x2+1xx-1=tan-1-7⇒tan-12×2-x+1-x+1=tan-1-7
So, we get
2×2-x+1-x+1=-7⇒2×2-x+1=7x-7⇒2×2-8x+8=0⇒x2-4x+4=0⇒x-22=0⇒x=2
Q26.
Answer :
(b) 32
We know that sin-1x+cos-1x=π2.
∴ π2-cos-1x-cos-1x=π6⇒-2cos-1x=π6-π2⇒-2cos-1x=-π3⇒cos-1x=π6⇒x=cosπ6⇒x=32
Q27.
Answer :
(b) π4
We know
tan-1x+tan-1y=tan-1x+y1-xy
∴ tan-1ab+c+tan-1bc+a=tan-1ab+c+bc+a1-ab+c×bc+a =tan-1ac+a2+b2+bcb+cc+aac+c2+bcb+cc+a
=tan-1ac+c2+bcac+c2+bc ∵a2+b2=c2 =tan-11=tan-1tanπ4=π4
Q28.
Answer :
(c) 122
Let sin-1638=y
Then,
siny=638cosy=1-sin2y=1-6364=18
Now, we have
sin14sin-1638=sin14y=1-cosy22 ∵cos2x=1-2sin2x=1-1+cosy22 ∵cos2x=2cos2x-1=1-1+1822=1-9162=1-342=18=122
Q29.
Answer :
(a) 7
Let 2cot-13=y
Then, coty2=3
cotπ4-2cot-13=cotπ4-y=cotπ4coty+1coty-cotπ4=coty+1coty-1 =cot2y2-12coty2+1cot2y2-12coty2-1=cot2y2+2coty2-1cot2y2-2coty2-1=9+6-19-6-1=7
Q30.
Answer :
(c) ±π6
We have,tan-1cotθ=2θ⇒tan2θ=cotθ⇒2tanθ1-tan2θ=1tanθ⇒2tan2θ=1-tan2θ⇒3tan2θ=1⇒tan2θ=13⇒tanθ=±13∴ θ=±π6