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Page No 7.19:

Question 1:

If sin A=45and cos B=513, where 0 < A, B<π2, find the values of the following:

(i) sin (A + B)
(ii) cos (A + B)
(iii) sin (A − B)
(iv) cos (A − B)

Answer:

Given:       sinA = 45 and cosB = 513We know that        cosA = 1 - sin2A       and   sinB = 1 - cos2B          ,  where 0 < A , B < π2  cosA = 1 - 452      and    sinB = 1 - 5132  cosA =1 - 1625        and    sinB = 1 - 25169  cosA =925                and    sinB = 144169  cosA =35                      and     sinB = 1213

Now,

(i) sinA+B = sinA cosB + cosA sinB                      =45×513 + 35×1213                     =2065 + 3665                     =5665

(ii) cosA+B = cosA cosB - sinA sinB                        =35×513 - 45×1213                        =1565 - 4855                        =-3365


(iii) sinA-B = sinA cosB - cosA sinB                         =45×513 - 35×1213                        =2065 - 3665                        =-1665


(iii) sinA-B=sinA cosB-cosA sinB                        =45×513-35×1213                        =2065-3665                        =-1665

iv cosA-B = cosA cosB + sinA sinB                         =35×513 + 45×1213                         =1565 + 4865                         =6365

Page No 7.19:

Question 2:

(a) If sin A=1213and sin B=45, where π2< A < π and 0 < B < π2, find the following:

(i) sin (A + B)
(ii) cos (A + B)

(b) If sin A=35, cos B=-1213, where A and B both lie in second quadrant, find the value of sin (A + B).

Answer:


a Given:sinA = 1213 and sinB = 45When, π2 < A < π and 0 < B < π2,cosA = -1 - sin2A    and cosB = 1 - sin2B As cosine function is negative in second qudrant and positive in first quadrant

 cosA =- 1 - 12132   and  cosB = 1 - 452 cosA =- 1 - 144169   and  cosB = 1 - 1625 cosA =- 25169   and  cosB = 925 cosA = -513   and  cosB = 35Now,


 (i) sinA+B = sinA cosB + cosA sinB                      =1213×35 + -513×45                      =3665 + -2065                      =1665ii cosA+B = cosA cosB - sinA sinB                        =-513×35 - 1213×45                        =-1565 - 4865                       =-6365

b Given:     sinA =35  and  cosB = -1213and that A and B both lie in second qudrant.We know that in second quadrant sine function  is positive and cosine function is negative.Therefore,     cosA =- 1 - sin2A         and      sinB = 1 - cos2B  cosA =- 1 - 352       and      sinB = 1 - -12132  cosA =- 1 -925          and      sinB = 1 - 144169 cosA =- 1625                and      sinB = 2569 cosA = -45                       and      sinB = 513Now, sinA+B = sinA cosB + cosA sinB                 =35×-1213 + -45×513                 =-3665  - 2065                 =-5665

Page No 7.19:

Question 3:

If cos A=-2425and cos B=35, where π < A < 3π2and3π2< B < 2π, find the following:

(i) sin (A + B)
(ii) cos (A + B)

Answer:

Given:cosA =-2425      and        cosB = 35and π<A<3π2    and   3π2<B<2π.That is, A is in third quadrant and B is in fourth qudrant.We know that sine function is negative in third and fourth quadrants.Therefore,sinA =- 1 - cos2A              and          sinB =- 1 - cos2BsinA = 1 - -24252       and          sinB = -1 - 352sinA =- 1 - 576625         and          sinB = -1 - 925sinA =- 49625                 and          sinB =-1625sinA = -725                          and          sinB = -45

Now,i sinA+B = sinA cosB + cosA sinB                       =-725×35 + -2425×-45                       =-21125+96125                       =75125                      
                        =35

ii cosA+B = cosA cosB - sinA sinB                        =-2425×35 - -725×-45                        =-72125 - 28125                        =-100125                        =-45

Page No 7.19:

Question 4:

If tan A=34, cos B=941, where π < A < 3π2and 0 < B < π2, find tan (A + B).

Answer:

Given:tanA = 34 and cosB = 941Here, π< A < 3π2 and 0 < B < π2.That is, A is in third quadrant and B is in first qudrant.We know that tan function is positive in first and third quadrants,and in the first quadrant, sine function is also positive.Therefore, sinB = 1 - cos2B                             =1 - 9412                             =1 - 811681                             =16001681                             =4041And tanB = sinBcosB                 =4041941=409Therefore, tanA+B = tanA + tanB1-tanA tanB                                     =34+4091-34×409                                     =18736-8436                                     =-18784

Page No 7.19:

Question 5:

If sin A=12, cos B=1213, where π2< A < π and 3π2 < B < 2π, find tan (AB).

Answer:

Given:sinA = 12      and       cosB = 1213Here, π2 < A < π   and  3π2< B < 2π.That is, A is in the second quadrant and B is in the fourth quadrant.We know that in the second quadrant, sine function is positive and cosine and tan functions are negative.In the fourth quadrant, sine and tan functions are negative and cosine function is positive. Therefore,cosA =- 1 - sin2A =- 1 - 122 = -1-14 = -34  = -32tanA = sinAcosA=12-32 = -13sinB =- 1 - cos2B =- 1 - 12132 =- 1 - 144169 =- 25169 = -513tanB = sinBcosB = -5131213 = -512Now, tanA-B = tanA -tanB1 + tanA tanB                            =-13--5121+-13×-512                            =-12+53123123+5123=53-125+123

Page No 7.19:

Question 6:

If sin A=12, cos B=32, where π2 < A < π and 0 < B < π2, find the following:

(i) tan (A + B)
(ii) tan (AB)

Answer:

Given: sinA = 12 and cosB = 32Here, π2<A<π and 0<B<π2.That is, A is in the second quadrant and B is in the first quadrant.We know that in the second quadrant, sine function is positive and cosine and tan functions are negativeIn the first quadrant, all T-functions are positive.Therefore,cosA = -1-sin2A = -1-122 =-1-14 =-34 =-32tanA = sinAcosA = 12-32 = -13sinB = 1-cos2A = 1-322 = 1-34 = 14 = 12tanB = sinBcosB = 1232 = 13

Now,i tanA+B = tanA + tanB1 - tanA tanB                       =-13+131--13×13                      =01+13= 0

ii tanA-B = tanA - tanB1+tanA tanB                        =-13-131+-13×13                       =-231-13                       =-2323                       =-3

Page No 7.19:

Question 7:

Evaluate the following:

(i) sin 78° cos 18° − cos 78° sin 18°
(ii) cos 47° cos 13° − sin 47° sin 13°
(iii) sin 36° cos 9° + cos 36° sin 9°
(iv) cos 80° cos 20° + sin 80° sin 20°

Answer:

i sin78°cos18° - cos 78° sin 18°    = sin78°-18°                                   Using sinA cosB - cosA sinB = sinA-B   = sin60° = 32ii cos47°cos13° -sin47° sin13°    =cos47°+13°                                   Using cosA cosB - sinA sinB = cosA+B     = cos60° = 12iii  sin36°cos9° + cos 36° sin9°    = sin36°+9°                                   Using sinA cosB+ cosA sinB = sinA+B   = sin45° = 12iv cos80°cos20° +sin80° sin20°    =cos80°-20°                                   Using cosA cosB + sinA sinB = cosA-B     = cos60° = 12

Page No 7.19:

Question 8:

If cos A=-1213and cot B=247, where A lies in the second quadrant and B in the third quadrant, find the values of the following:

(i) sin (A + B)
(ii) cos (A + B)
(iii) tan (A + B)

Answer:

Given:cosA = -1213 and cotB = 247A lies in the second quadrant and B lies in the third quadrant.We know that sine function is positive in the second quadrant and in the third quadrant, both sine and cosine functions are negative.Therefore,sinA = 1-cos2A = 1--12132 = 1-144169 = 25169 = 513sinB = -11+cot2B = -11+2472 = -11+57649=-162549=-725cosB = -1-sin2B = -1--7252 =- 1-49625 =- 576625=-2425Now,i sinA+B = sinA cosB + cosA + sinB                       = 513×-2425 + -1213×-725                     =-120325+84325                     =-36325ii cosA+B=cosA cosB - sinA sinB                        =-1213×-2425 - 513×-725                        =288325 + 35325                       =323325iii tanA+B=sinA+BcosA+B =-36325323325 =- 36323

Page No 7.19:

Question 9:

Prove that: 7π12+cosπ12=sin5π12-sinπ12

Answer:

105°=7π12, 15°=π12, 75°=5π12, 15°=π12
LHS = cos105o + cos15o
        = cos(90o + 15o) + cos(90o - 75o)
        = - sin 15o + sin 75o                      [As cos(90o+A) = - sin A and cos(90o - B) = sin B]
        = sin 75o - sin 15o
        = RHS
Hence proved.

Page No 7.19:

Question 10:

Prove that tan A+tan Btan A- tan B=sin A+Bsin A-B .

Answer:

LHS = tanA + tanBtanA - tanB             =sinAcosA+sinBcosBsinAcosA-sinBcosB             =sinA cosB+cosAsinBcosA cosBsinA cosB - cosA sinBcosA cosB            =sinA cosB + cosA sinBsinA cosB -cosA sinB            =sinA+BsinA-B             =RHSHence proved.

Page No 7.19:

Question 11:

Prove that
(i) cos 11°+sin 11°cos 11°-sin 11°=tan 56°.

(ii) cos 9°+sin 9°cos 9°-sin 9°=tan 54°
(ii) cos 8°-sin 8°cos 8°+sin 8°=tan 37°

Answer:

(i)
LHS= cos11° + sin11°cos11° - sin11°             =cos11°cos11°+sin11°cos11°cos11°cos11° - sin11°cos11°           Dividing numerator and denominator by cos11°            =1+tan11°1-tan11°            =1+tan11°1-1×tan11°            =tan45° + tan11°1 - tan45° tan11°           As tan45° = 1            = tan45°+11°                    As tanA + tanB1 - tanA tanB = tanA+B            = tan56°            =  RHSHence proved.

i LHS = cos9° + sin9°cos9° - sin9°                =cos9°cos9° + sin9°cos9°cos9°cos9° - sin9°cos9°                Dividing the numerator and denominator by cos9                =1 + tan9°1 -  tan9°                =1 + tan9°1 + 1× tan9°                =tan45°+ tan9°1 - tan45°× tan9°                As tan45° = 1                = tan45° + 9°                As  tanA+tanB1 - tanA tanB = tanA+B                = tan54°                = RHSHence proved.

ii LHS = cos8° - sin8°cos8° + sin8°                 =cos8°cos8° - sin8°cos8°cos8cos8 +sin8cos8             Dividing numeraor and denominator by cos8°                  =1-tan8°1+tan8°                  =1- tan8°1+1×tan8°                  =tan45° - tan8°1 + tan45° tan8°              As tan 45° = 1                  =tan45°-8°                       As tanA-tanB1+tanA tanB=tanA+B                  =tan37°                  =RHSHence proved.

Page No 7.19:

Question 12:

Prove that:
(i) sinπ3-xcosπ6+x+cosπ3-xsinπ6+x=1
(ii) sin4π9+7cosπ9+7-cos4π9+7sinπ9+7=32
(iii) sin3π8-5cosπ8+5+cos3π8-5sinπ8+5=1

Answer:

(i) π3=60°, π6=30°
LHS = sin60°-x cos30°+x +cos60°-x sin30°+x             =sin60°-x+30°+x          Using the formula sinA cosB + cosA sinB =  sinA+B                           and taking A =60°-x and B =30°+x                      =sin90°           =1            = RHSHence proved.

(ii)
sin4π9+7cosπ9+7-cos4π9+7sinπ9+7=sin4π9+7-π9+7                  sinAcosB-cosAsinB=sinA-B=sin3π9=sinπ3=32

(iii)
sin3π8-5cosπ8+5+cos3π8-5sinπ8+5=sin3π8-5+π8+5                sinAcosB+cosAsinB=sinA+B=sin4π8=sinπ2=1

Page No 7.19:

Question 13:

Prove that tan 69°+tan 66°1-tan 69° tan 66°=-1.

Answer:

LHS = tan69°+tan66°1-tan69°tan66°             =tan69°+66°             Using the formula tanA+tanB1-tanAtanB=tanA+B             =tan135°             =tan180°-45°             =-tan45°            tan180-A = -tanA             = -1              = RHSHence proved. 



Page No 7.20:

Question 14:

(i) If tan A=56and tan B=111, prove that A+B=π4.
(ii) If tan A=mm-1and tan B=12m-1, then prove that A-B=π4.

Answer:

(i)
We have:tanA = 56 and tanB = 111Therefore, tanA+B = tanA+tanB1-tanA tanB             tanA+B = tanA +tanB1 - tanA tanB            tanA+B = 56+1111 - 56×111            tanA+B =61666166            tanA+B =1            tanA+B =tanπ4Therefore, A + B = π4.Hence proved. 

(ii)
We know thattan(A-B)=tan A-tan B1+tan Atan B                  =mm-1-12m-11+m(m-1)(2m-1)                  =2m2-m-m+12m2-m-2m+1+m                  =2m2-2m+12m2-2m+1                  =1A-B =tan-1(1) A-B =π4

Page No 7.20:

Question 15:

Prove that:
(i) cos245°-sin215°=34
(ii) sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.

Answer:

(i)
cos245°-sin215°=cos45°+15°cos45°-15°                  cos2X-sin2Y=cosX+YcosX-Y=cos60°cos30°=12×32=34

Hence proved.

(ii)
LHS = sin2n+1A - sin2nA             =sinn+1A+nA sinn+1A-nA      Using the formula sin2X - sin2Y = sinX+Y sinX-Y and taking X = n+1A and Y = nA            =sinn+1+nA sin n+1-nA            =sin2n +1A sinA            = RHSHence proved.

Page No 7.20:

Question 16:

Prove that:

(i) sin A+B+sin A-Bcos A+B+cos A-B=tan A
(ii) sin A-Bcos A cos B+sin B -Ccos B cos C+sin C-Acos C cos A=0
(iii) sin A-Bsin A sin B+sin B-Csin B sin C+sin C-Asin C sin A=0
(iv) sin2 B = sin2 A + sin2 (AB) − 2 sin A cos B sin (A B)
(v) cos2 A + cos2 B − 2 cos A cos B cos (A + B) = sin2 (A + B)
(vi) tan A+Bcot A-B=tan2A-tan2 B1-tan2 A tan2 B

Answer:

i LHS= sinA+B+sinA-BcosA+B +cosA-B                 =sinA cosB +cosA sinB + sinA cosB - cosA sinBcosA cosB -sinA sinB +cosA cosB +sinA sinB                 =2sinA cosB2cosA cosB                 =sinAcosA                 =tanA                 = RHSHence proved.

ii LHS = sinA-BcosA cosB+sinB-CcosB cosC + sinC-AcosC cosA                  =sinA cosB - cosA sinBcosA cosB + sinB cosC -cosB sinCcosB cosC +sinC cosA-cosC sinAcosC cosA                  =sinA cosBcosA cosB - cosA sinBcosA cosB +sinB cosCcosB cosC - cosB sinCcosB cosC +sinC cosAcosC cosA -cosC sinAcosC cosA                  =sinAcosA - sinBcosB +sinBcosB - sinCcosC +sinCcosC-sinAcosA                  =tanA - tanB + tanB -tanC +tanC -tanA                  = 0                  =RHSHence proved.

iii LHS = sinA-BsinA sinB +sinB-CsinB sinC + sinC-AsinC sinA                  =sinA cosB -cosA sinBsinA sinB + sinB cosC -cosB sinCsinB sinC+sinC cosA -cosC sinAsinC sinA                  =sinA cosBsinA sinB - cosA sinBsinA sinB + sinB cosCsinB sinC - cosB sinCsinB sinC + sinC cosAsinC sinA - cosC sinAsinC sinA                  =cosBsinB - cosAsinA + cosCsinC-cosBsinB +cosAsinA - cosCsinC                  = cotB - cotA + cotC -cotB+cotA-cotC                  =0                  =RHSHence proved.

iv RHS= sin2A + sin2A-B -2sinA cosB sinA-B                      = sin2A + sinA-B sinA-B -2sinA cosB                      = sin2A + sinA-B sinA cosB - cosA sinB - 2sinA cosB                      = sin2A + sinA-B -sinA cosB - cosA sinB                      = sin2A - sinA-B sinA cosB + cosA sinB                      = sin2A - sinA-B sinA+B                      = sin2A - sin2A - sin2B                      = sin2A - sin2A + sin2B                      =  sin2B                      = LHSHence proved.

v LHS = cos2A + cos2B -2cosA cosB cosA+B                    = cos2A + 1 - sin2B - 2cosA cosB cosA+B                    =1 + cos2A - sin2B - 2cosA cosB cosA+B                   =1 + cos2A - sin2B - 2cosA cosB cosA+B                   =1 + cosA+BcosA-B - 2cosA cosB cosA+B                   =1 + cosA+BcosA-B - 2cosA cosB                   =1 + cosA+BcosA cosB + sinA sinB - 2cosA cosB                    =1 + cosA+B-cosA cosB + sinA sinB                   =1 - cosA+BcosA cosB - sinA sinB                   =1 - cosA+BcosA+B                   =1 - cos2A+B                   = sin2A+B                    = RHSHence proved.

vi LHS =tanA+BcotA-B                    =tanA+B1tanA-B                    = tanA+B × tanA-B                    = tanA +tanB1-tanA tanB×tanA -tanB1+tanAtanB                    =tanA+tanBtanA -tanB1-tanA tanB1 + tanA tanB                    =tanA2-tanB212-tanA tanB2                    =tan2A -tan2B1 - tan2A tan2B                    = RHSHence proved.

Page No 7.20:

Question 17:

Prove that:
(i) tan 8x − tan 6x − tan 2x = tan 8x tan 6x tan 2x
(ii) tanπ12+tanπ6+tanπ12tanπ6=1
(iii) tan 36° + tan 9° + tan 36° tan 9° = 1
(iv) tan 13x − tan 9x − tan 4x = tan 13x tan 9x tan 4x

Answer:

i We know that 8x = 6x + 2xTherefore,    tan8x = tan6x + 2x tan8x =tan6x+tan2x1 - tan6x tan2xtan8x - tan8x tan6x tan2x = tan6x + tan2xtan8x - tan6x - tan2x =tan8x tan6x tan2xHence proved.

ii π12=15°, π6=30°We know that 45° = 15° + 30°Therefore,tan45° = tan15°+30°1 = tan15° + tan30°1 - tan15° tan30° 1 - tan15° tan30° = tan15° + tan30°  1 =  tan15° + tan30° + tan15° tan30° tan15° + tan30° + tan15° tan30°= 1Hence proved.

iii We know that 36° + 9° = 45°Therefore,   tan36° + 9° = tan45°tan36° + tan9°1 - tan36° tan9° = 1tan36° + tan9° = 1 - tan36° tan9°tan36° + tan9° + tan36° tan9° = 1Hence proved.

iv We know that 13x = 9x + 4xTherefore,    tan13x = tan9x + 4xtan13x = tan9x + tan4x1 - tan9x tan4xtan13x - tan13x tan9x tan 4x = tan9x + tan4xtan13x - tan9x - tan4x = tan13x tan9x tan4x Hence proved.

Page No 7.20:

Question 18:

Prove that: tan2 2x-tan2 x1-tan2 2x tan2 x=tan 3x tan x

Answer:

LHS =tan22x-tan2x1-tan22x tan2x=(tan2x+tanx)(tan2x-tanx)1-tan22x tan2x        Using A2-B2=A+BA-B tan3x=tan(2x+x) and tanx=tan(2x-x).=tan3x1-tan2x tanx×tanx1+tan2x tanx1-tan22x tan2x             tan 2x+tan x=tan3x1-tan2x tanx & tan 2x -tanx=tanx1+tan2x tanx=tan3x tanx (1-tan22x tan2x)1-tan22x tan2x=tan3x tanx  

=RHS Hence proved. 

Page No 7.20:

Question 19:

Prove that sin2 (n + 1) A − sin2 nA = sin (2n + 1) A sin A.

Answer:

LHS = sin2n+1A - sin2nA             =sinn+1A+nA sinn+1A-nA      Using the formula sin2X - sin2Y = sinX+Y sinX-Y and taking X = n+1A and Y = nA            =sinn+1+nA sin n+1-nA            =sin2n +1A sinA            = RHSHence proved.

Page No 7.20:

Question 20:

If tan A = x tan B, prove that sin A-Bsin A+B=x-1x+1.

Answer:

LHS = sin(A-B)sin(A+B)       = sin A cos B - cos A sin Bsin A cos B + cos A sin BDividing numerator and denominator by cos A cos B:        tan A -tan Btan A + tan B        =xtan B-tan Bxtan B+tan B             (Since tan A = x tan B )         =tan B x-1tan B x+1         =  x-1x+1    = RHSHence proved.

Page No 7.20:

Question 21:

If tan (A + B) = x and tan (AB) = y, find the values of tan 2A and tan 2B.

Answer:

tan(2A)=tan(A+A)             =tan(A+B+A-B)            =tan(A+B)+tan(A-B)1-tan(A+B)tan(A-B)           =x+y1-xy

tan 2B= tan B+B            =tan B+A+B-A            = tan A+B+tan B-A1-tanA+BtanB-A            =tanA+B-tanA-B1+tanA+BtanA-B        tan-θ=-tan θ           =x-y1+xy

Page No 7.20:

Question 22:

If cos A + sin B = m and sin A + cos B = n, prove that 2 sin (A + B) = m2 + n2 − 2.

Answer:

RHS= m2+n2-2= cos A +sin B2+sin A+cos B 2-2=cos2A+sin2B+2cosAsinB+sin2 A+cos 2 B +2sinAcos B-2=1+1+2cosAsinB+2sinAcos B-2=2cos AsinB+sinA cos B=2sinA+B=LHSHence proved.

Page No 7.20:

Question 23:

If tan A + tan B = a and cot A + cot B = b, prove that cot (A + B) 1a-1b.

Answer:

Given:

cot A + cot B = b1tanA+1tanB=btanA+tanBtanAtanB=bNow,RHS = 1a-1b          = 1tan A +tan B-tan A tan Btan A +tan B        = 1-tan A tan Btan A + tan B         =  cot (A+B)         = LHSHence proved.

Page No 7.20:

Question 24:

If x lies in the first quadrant and cos x=817, then prove that:
cos π6+x+cos π4-x+cos 2π3-x=3-12+122317

Answer:

Given: 0<x<π2Now, sin x = 1-cos2x = 1-64289=1517LHS = cosπ6+x +cosπ4-x+cos2π3-x        =  cos(30+x) +cos(45-x)+cos(120-x)        =cos 30° cos x -sin30° sin x +cos 45° cos x+sin 45° sin x + cos120° cos x+sin120° sin x           Using formulas of cos(A+B) and cos(A-B)        = cos x(cos 30°+cos  45°+cos120) +sin x(-sin 30° +sin 45° +sin 120°)       = 81732+12-12 +1517-12+12+32             =8173-12+12+15173-12+12     =23173-12+12        = RHSHence proved.

Page No 7.20:

Question 25:

If tan x + tan x+π3+tan x+2π3=3, then prove that 3 tan x- tan3 x1-3 tan2 x=1.

Answer:

Given:tan x +tanx+π3+tanx+2π3 = 3tanx+tan x+tanπ31-tan x tan π3+tan x+tan2π31-tan x tan2π3=3tanx+tan x+31-3tan x +tan x-31+3tan x =3              tan120°=-3tan x(1-3tan2x)+tan x+3+3tan2x+3tan x+tan x-3-3tan2x+3tan x1-3tan2x=3 9tan x-3tan3x1-3tan2x=33tan x-tan3x1-3tan2x=1Hence proved.



Page No 7.21:

Question 26:

If sin (α + β) = 1 and sin (α − β)=12, where 0 ≤ α, βπ2, then find the values of tan (α + 2β) and tan (2α + β).

Answer:

Given:sin (α+β)=1 and sin (α-β)=12α+β=90°         ...(1) and α-β=30°     ...(2) By adding eq (1) and eq (2) we get:  2α=120°α=60°By subtracting  eq (2) from eq (1), we get: 2β=60° β=30°Therefore, tan(α+2β)=tan 60°+2×30°=tan 120°=-3tan(2α+β)=tan 2×60°+30°=tan 150° = -13

Page No 7.21:

Question 27:

If α, β are two different values of x lying between 0 and 2π, which satisfy the equation 6 cos x + 8 sin x = 9, find the value of sin (α + β).

Answer:

Given:6 cosx+8 sinx=96 cosx=9-8 sinx36cos2x=(9-8 sinx)2361-sin2x=81+64sin2x -144sinx100sin2x -144sinx+45 =0Now, α and β are the roots of the given equation; therefore, cos α and cos β are the roots of the above equation.sinα sinβ=45100          (Product of roots of a quadratic equation ax2+bx+c=0 is ca.)Again, 6 cosx+8 sinx=98 sinx=9-6 cosx64sin2x=(9-6 cosx)264(1-cos2x)=81+36cos2x-108cos x100cos2x-108cosx+17=0Now, α and β are the roots of the given equation; therefore, sinα and sinβ are the roots of the above equation.Therefore, cosα cosβ=17100Hence, cos(α+β)=cosα cosβ-sinα sinβ                               =17100-45100                              =-28100                              =-725

sin α+β =1-cos2α+β                  =1--7252                      =576625                  =2425

Page No 7.21:

Question 28:

If sin α + sin β = a and cos α + cos β = b, show that

(i) sin α+β=2aba2+b2
(ii) cos α+β=b2-a2b2+a2

Answer:

(i)
a2+b2=sinα+sinβ2+(cosα+cosβ)2a2+b2=sin2α+sin2β+2sinαsinβ+cos2α+cos2β+2cosαcosβa2+b2=sin2α+cos2α+sin2β+cos2β+2sinαsinβ+cosαcosβa2+b2=2+2 cos(α-β)                     ...(1)

Now,

b2-a2=(cosα+cosβ)2-sinα+sinβ2b2-a2 = cos2α+cos2β -sin2α-sin2β +2cosαcosβ-2sinαsinβb2-a2  =(cos2α-sin2β)+(cos2β-sin2α)-2cos(α+β) b2-a2 =2cos(α+β)cos(α-β)+2cos(α-β) b2-a2 =cos(α+β)(2+2 cos(α-β))   ...(2)

From (1) and (2), we have

b2-a2 =cos(α+β)a2+b2     b2-a2a2+b2 =cos(α+β)

sinα+β =1-cos2(α+β)sinα+β=1-b2-a2b2+a22 =b4+a4-b4-a4+4a2b2b2+a22  sinα+β=2aba2+b2

(ii)

a2+b2=sinα+sinβ2+(cosα+cosβ)2             =sin2α+sin2β+cos2α+cos2β+2sinαsinβ+2cosαcosβ             =2+2 cos(α-β)b2-a2=(cosα+cosβ)2-sinα+sinβ2b2-a2 = cos2α+cos2β -sin2α-sin2β +2cosαcosβ-2sinαsinβb2-a2  =(cos2α-sin2β)+(cos2β-sin2α)-2cos(α+β) b2-a2 =2cos(α+β)cos(α-β)+2cos(α-β) b2-a2 =cos(α+β)(2+2 cos(α-β))     b2-a2 =cos(α+β)a2+b2     b2-a2a2+b2 =cos(α+β)

Page No 7.21:

Question 29:

Prove that:

(i) 1sin x-a sin x-b=cot x-a-cot x-bsin a-b
(ii) 1sin x-a cos x-b=cot x-a+tan x-bcos a-b
(iii) 1cos x-a cos a-b=tan x-b-tan x-asin a-b

Answer:

(i) RHS= cotx-a -cot(x-b)sin(a-b)         = cos(x-a)sin(x-a)-cos(x-b)sin(x-b)sin(a-b)         = sin(x-b) cos(x-a) -sin(x-a) cos(x-b)sin(x-a) sin(x-b) sin(a-b)          =sin(x-b -x +a)sin(x-a) sin(x-b) sin(a-b)         = sin(a-b)sin(x-a) sin(x-b) sin(a-b)           = 1sin(x-a)sin(x-b)           =LHSHence proved.  


(ii) RHS = cot(x-a) + tan(x-b)cos(a-b)           = cos(x-a)sin(x-a)+sin(x-b)cos(x-b)cos(a-b)          = cos(x-b) cos(x-a) + sin(x-a) sin(x-b)cos(a-b) sin(x-a) cos(x-b)          =cos(x-b-x+a)cos(a-b) sin(x-a) cos(x-b)        ( Using   cos(A-B) = cos A cosb B+sin A sin B)          =cos(a-b)cos(a-b) sin(x-a) cos(x-b)=    1 sin(x-a) cos(x-b)   = RHSHence proved. 


(iii) RHS = tan(x-b) -tan(x-a)sin(a-b)         =sin(x-b)cos(x-b)-sin(x-a)cos(x-a)sin(a-b)        = sin(x-b) cos(x-a) -sin(x-a) cos(x-b)sin(a-b) cos(x-a) cos(x-b)        = sin(x-b-x+a)sin(a-b) cos(x-a) cos(x-b)        ( Using sin(A-B) = sin Acos B -cos Asin B)         =sin(a-b)sin(a-b) cos(x-a) cos(x-b)        = 1cos(x-a) cos(x-b)        =LHSHence proved. 

Page No 7.21:

Question 30:

If sin α sin β − cos α cos β + 1 = 0, prove that 1 + cot α tan β = 0.

Answer:

Given:
sinα sinβ − cosα cosβ + 1 = 0-(cosα cosβ-sinα sinβ) +1 = 0-cos(α+β) +1 = 0cos(α+β) = 1Therefore, sin(α+β) = 0          ....(1) (Since sinθ = 1-cos2θ )      Hence ,1+cotα tanβ  = 1 +cosα sinβ sinα cosβ                          = sinαcosβ +cosαsinβsinα cosβ                         = sin(α+β)sinαcosβ                         = 0             ...From eq (1) Hence proved.

Page No 7.21:

Question 31:

If tan α = x +1, tan β = x − 1, show that 2 cot (α − β) = x2.

Answer:

LHS= 2cot(α-β)       =2(1+tanαtanβ)tanα-tanβ

        =2+2x+1x-1x+1-x+1        =2+2x2-22         =2x22          =x2          =RHS Hence proved.

Page No 7.21:

Question 32:

If angle θ is divided into two parts such that the tangents of one part is λ times the tangent of other, and ϕ is their difference, then show that sinθ=λ+1λ-1sinϕ.                                                                                                                                                          [NCERT EXEMPLER]

Answer:

Let α and β be the two parts of angle θ. Then,

θ=α+β and ϕ=α-β               (Given)

Now,

tanα=λ tanβ                Giventanαtanβ=λ1

Applying componendo and dividendo, we get

tanα+tanβtanα-tanβ=λ+1λ-1sinαcosα+sinβcosβsinαcosα-sinβcosβ=λ+1λ-1sinα cosβ+cosα sinβcosα cosβsinα cosβ-cosα sinβcosα cosβ=λ+1λ-1sinα+βsinα-β=λ+1λ-1

sinθsinϕ=λ+1λ-1                    θ=α+β and ϕ=α-βsinθ=λ+1λ-1sinϕ                           

Page No 7.21:

Question 33:

If tanθ=sinα-cosαsinα+cosα, then show that sinα+cosα=2cosθ.                              [NCERT EXEMPLER]

Answer:

tanθ=sinα-cosαsinα+cosα

Dividing numerator and denominator on the RHS by cosα, we get

tanθ=sinαcosα-1sinαcosα+1tanθ=tanα-tanπ41+tanα tanπ4tanθ=tanα-π4θ=α-π4Or α=π4+θ

Now,

sinα+cosα=sinπ4+θ+cosπ4+θ=sinπ4cosθ+cosπ4sinθ+cosπ4cosθ-sinπ4sinθ=12cosθ+12sinθ+12cosθ-12sinθ=22cosθ=2cosθ

 sinα+cosα=2cosθ

Page No 7.21:

Question 34:

If α and β are two solutions of the equation a tan x + b sec x = c, then find the values of sin (α + β) and cos (α + β).

Answer:

a tan x+b sec x=cc-a tan x=b sec xc-a tan x2=b sec x2c2+a2 tan2 x-2ac tan x=b2 sec2 xc2+a2 tan2 x-2ac tan x=b21+tan2 xa2-b2 tan2 x-2ac tan x+c2-b2=0This is a quadratic in tan x.It has two solutions tan α and tan β.tan α+tan β=2aca2-b2tan α×tan β=c2-b2a2-b2Therefore, tanα+β=tan α+tan β1-tan αtan β                                      =2aca2-b21-c2-b2a2-b2                                      =2aca2-c2Hence, sinα+β=2aca2+c2 and cosα+β=a2-c2a2+c2.



Page No 7.26:

Question 1:

Find the maximum and minimum values of each of the following trigonometrical expressions:
(i) 12 sin x − 5 cos x
(ii) 12 cos x + 5 sin x + 4
(iii) 5 cos x+3 sin π6-x+4
(iv) sin x − cos x + 1

Answer:

(i)
Let fx =12 sin x - 5 cosxWe know that-122+(-5)212 sin x - 5 cosx122+(-5)2-144+2512 sin x - 5 cosx144+25-1312 sinx - 5 cosx13Hence the maximum and minumun values of fx are 13 and -13, respectively .

(ii)
Let f(x)=12 cosx +5 sinx +4We know that-122+5212 cosx +5 sinx 122+52             for all x-16912 cosx +5 sinx 169-1312 cosx +5 sinx 13-912 cosx +5 sinx +417Hence, the maximum and minimum vaues of fx are  17 and -9, respectively.

(iii)
Let fx=5 cosx +3 sinπ6-x +4Now  fx = 5cosx+3sin30°cosx -cos30°sinx+4                =5cosx +32cosx -332sinx +4                 =132cosx-332sinx +4We know that-1322+-3322132cosx-332sinx1322+-3322          for all xTherefore,-169+274 132cosx-332sinx 169+274-142+4132cosx-332sinx +4142+4-3132cosx-332sinx +411Hence, maximum and minimun values of fx are 11 and -3, respectively .

(iv)
Let fx =sinx-cosx+1We know that-12+(-1)2sinx-cosx12+(-1)2     for all x-2sinx-cosx2-2+1sinx-cosx +12+1Hence maximum and minimum values of f(x) are  1+2  and 1-2 , respectively .

Page No 7.26:

Question 2:

Reduce each of the following expressions to the sine and cosine of a single expression:
(i) 3 sin x-cos x
(ii) cos x − sin x
(iii) 24 cos x + 7 sin x

Answer:

(i) Let fx =3 sinx - cosxDividing and multiplying by3+1 , i.e. by 2, we get: fx =232 sinx -12 cosxf(x)= 2cosπ6sinx-sinπ6cosxf(x) = 2sinx-π6Again, fx =232 sinx -12 cosx fx =2sinπ3 sinx -cosπ3 cosxfx =-2cosπ3+x

(ii) Let fx=cosx-sinxDividing and multiplying by 12+12, i.e. by2, we get : fx=212cosx-12sinxfx=2(cos45°cosx-sin45°sinx)     fx=2cosπ4+xAgain,     fx=212cosx-12sinxfx=2(sin45°cosx-cos45°sinx)f(x) =2 sinπ4-x

(iii) Let f(x) =24 cosx+7sinxDividing and multiplying by 242+72, i.e.by  25, we get: f(x) =252425 cosx +725sinxf(x) =25(sinα cosx+ cosα sinx),  where sinα =2425 and cosα =725f(x) =25 sin(α+x), where tanα = 247 .Again, f(x) =252425 cosx +725sinxf(x) =25(cosα cosx+ sinα sinx),  where cosα=2425, sinα= 725.f(x) =25 cos(α-x), where tanα =724.

Page No 7.26:

Question 3:

Show that sin 100° − sin 10° is positive.

Answer:

Let fθ = sin100° - sin10°Multiplying and dividing by 12+12   , i.e. by 2 , we get:              212sin100°-12sin10°             =2cos45°sin(90°+10°) -sin45°sin10°             =2cos45°cos10°-sin45°sin10°             =2 cos(45°+10°) =2co s55° , which is positive since cos is positive in the first quadrant.

Page No 7.26:

Question 4:

Prove that 23+3 sin x+23 cos x lies between -23+15 and 23+15.

Answer:

Let fx =(23+3) sinx +23cosxWe know that,-23+32+232fx23+32+232-12+9+123+12fx12+9+123+12-33+123 fx33+123Disclaimer : Instead of -23+15 and 23+15, it should be -33+123and  33+123.

Page No 7.26:

Question 1:

If α + β − γ = π and sin2 α +sin2 β − sin2 γ = λ sin α sin β cos γ, then write the value of λ.

Answer:

Given:γ=-π-(α+β)Also,λ=sin2α+sin2β-sin2-(π-(α+β)sinα sinβ cos(-(π-(α+β))          =sin2α+sin2β-(sin(α+β))2-(sinα sinβcos(α+β))        sin π-θ=sin θ and cosπ-θ=-cos θ   = sin2α+sin2β-sin2α cos2β-cos2α sin2β-2sinα sinβ cosα cosβ-(sinα sinβ cosα cosβ-sin2α sin2β)   =sin2α(1-cos2β)+sin2β(1-cos2α)-2sinα sinβ cosα cosβsin2α sin2β -sinα sinβ cosα cosβ  =2sin2α sin2β-2sinα sinβ cosα cosβsin2α sin2β-sinα sinβ cosα cosβ   =2

Page No 7.26:

Question 2:

If x cos θ = y cos θ+2π3=z cos θ+4π3, then write the value of 1x+1y+1z.

Answer:

Given:x cosθ =ycosθcos2π3-sinθ sin2π3=zcosθcos4π3-sinθ sin4π3xcosθ =y-12cosθ-32sinθ =z-12cosθ+32sinθ x=y2-1-3tanθ=z2-1+3tanθx=y2-1-3tanθz=y-1-3tanθ-1+3tanθNow,1x+1y+1z =2y-1-3tanθ+1y+-1+3tanθy-1-3tanθ                      =2+-1-3tanθ+-1+3tanθy-1-3tanθ                      =0

Page No 7.26:

Question 3:

Write the maximum and minimum values of 3 cos x + 4 sin x + 5.

Answer:

Let fx =3 cos x +4 sin x +5We know that-32+423 cosx +4 sinx32+42-53 cosx +4 sinx5-5+53 cosx +4 sinx+55+50f(x)10Hence, maximum and minimum vales of f(x) are  0 and 10 respectively .

Page No 7.26:

Question 4:

Write the maximum value of 12 sin x − 9 sin2 x.

Answer:

Let fx=12sinx -9 sin2x                =-9sin2x-12 sin x                = -3sin x2-2.3 sin x.2+22-4               =-3 sinx-22-4               =4-3 sinx-22Minimum value of  3 sinx-22 is 0.Therefore, maximum value of  4-3 sinx-22 would be 4.

Page No 7.26:

Question 5:

If 12 sin x − 9sin2 x attains its maximum value at x = α, then write the value of sin α.

Answer:

Let fx=12sinx -9 sin2x                =-9sin2x-12 sin x                = -3sin x2-2.3 sin x.2+22-4               =-3 sinx-22-4               =4-3 sinx-22Minimum value of  3 sinx-22 is 0.Therefore, maximum value of fx= 4-3 sinx-22 is  4.We are given that 12sinx -9 sin2x will attain its maximum value at x=α. 12sinα -9 sin2α =4-9sin2α+12sinα-4=09sin2α-12 sinα+4=09sin2α-6sinα-6sinα+4=03sinα3sinα-2-23sinα-2=03sinα-23sinα-2=0 sinα=23



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Question 6:

Write the interval in which the value of 5 cos x + 3 cos x+π3+3 lies.

Answer:

Let fx=5 cosx+3 cosx+π3+3            =5 cosx+3(cosx cos60° -sinx sin60°) +3             =5 cosx+32cosx-332sinx +3             =132cosx-332sinx +3We know that-1322+3322132cosx-332sinx1322+3322-1694+274132cosx-332sinx1694+274-142132cosx-332sinx142-7+3132cosx-332sinx+37+3Hence, f(x) lies in the interval -4,10.

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Question 7:

If tan (A + B) = p and tan (AB) = q, then write the value of tan 2B.

Answer:

tan 2B = tan(B+B)             =tan(A+B-(A-B))             = tan(A+B) - tan(A-B) 1+tan(A+B)tan(A-B)                =p-q1+pq                   tan(A+B)=p and tan(A-B)=q  

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Question 8:

If cos x-ycos x+y=mn, then write the value of tan x tan y.

Answer:

cos(x-y)cos(x+y)=mncosx cosy+ sinx sinycosx cosy - sinx siny=mn1+tanx tany1- tanx tany=mn                  Dividing numerator and denominator of LHS by cosx cosyn+ntanx tany =m-mtanx tanytanxtany(m+n)=m-ntanx tany =m-nm+n

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Question 9:

If a = b cos 2π3=c cos4π3, then write the value of ab + bc + ca.

Answer:

a=b cos120°=c cos 240°a =-12b=-12cTherefore,ab+bc+ca =-12b×b +b×b +b×-12b                      =-b2+b2                       =0

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Question 10:

If A + B = C, then write the value of tan A tan B tan C.

Answer:

tanA tanB tanC =tanA tanB tan(A+B)               Using A+B=C                            =tanA tanB×tanA+tanB1-tanA tanB                           =tan2AtanB+ tanA tan2B1-tanA tanB                           =tan2AtanB+ tanA tan2B+tanA+tanB-tanA-tanB1- tanA tanB                           =-tanA(1-tanAtanB)-tanB(1-tanAtanB)+tanA+tanB1-tanA tanB                           =-(1-tanAtanB)tanA+tanB+tanA+tanB1-tanA tanB                           =tanA+tanB1-tanA tanB-tanA-tanB                            =tan(A+B)-tanA-tanB                           =tanC-tanA-tanB

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Question 11:

If sin α − sin β = a and cos α + cos β = b, then write the value of cos (α + β).

Answer:

   cos(α+β) =cosαcosβ -sinα sinβ                  =2cosαcosβ -2sinα sinβ +2-22
                     =sin2α +cos2α+sin2β +cos2β+2cosαcosβ -2sinα sinβ-22
                     =(sinα-sinβ)2+(cosα+cosβ)2-22 
                     =a2+b2-22

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Question 12:

If tan α=11+2-xand tan β=11+2x+1, then write the value of α + β lying in the interval 0, π2.

Answer:

tan(α+β) =tanα+tanβ1-tanαtanβ                =11+2-x+11+2x+11-1(1+2-x)(1+2x+1)               =1+2x+1+1+2-x1+2x+1+2-x+2-x+x+1-1               =2+2x+1+2-x2+2x+1+2-x               =1Therefore, α+β =tan-1(1)=π4.

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Question 1:

The value of sin25π12-sin2π12 is
(a) 12
(b) 32
(c) 1
(d) 0

Answer:

(b)  32
5π12=75°, π12=15°

sin275°-sin215° =sin275°-cos275°           sin90°-θ=cosθ                               Now, sin75° =sin(45°+30°)                     =sin45°cos30°+cos45°sin30°                     =12×32+12×12                     =3+122cos75° =cos(45°+30°)              =cos45° cos30° -sin45° sin30°              =12×32-12×12              =3-122Hence ,sin275°-cos275°=3+1222-3-1222                         =3+1+23-3-1+238                        =438                        =32

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Question 2:

If A + B + C = π, then sec A (cos B cos C − sin B sin C) is equal to
(a) 0
(b) −1
(c) 1
(d) None of these

Answer:

(b)  −1
π = 180°
secAcosBcosC-sinBsinC=cosBcosπ-A+B-sinBsinπ-A+BcosA

We know that, cosπ-θ=-cosθ and sinπ-θ=sinθ,

secAcosBcosC-sinBsinC=cosBcosA+B-sinBsinA+BcosA

Now, using the identities cosA+B=cosAcosB-sinAsinB and sinA+B=sinAcosB+cosAsinB, we get

secAcosBcosC-sinBsinC=-cosAcosB2+cosBsinAsinB-sinBsinAcosB-sin2BcosAcosA

secAcosBcosC-sinBsinC=-cosAcos2B+sin2BcosAsecAcosBcosC-sinBsinC=-cosAcosA=-1

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Question 3:

tan 20° + tan 40° + 3 tan 20° tan 40° is equal to

(a) 34
(b) 32
(c) 3
(d) 1

Answer:

(c) 3

tan20°+tan40°+3tan20° tan40°=tan 60°(1-tan20°tan40°) +tan60°tan20°tan40°                      Using tan60°=tan20+tan401-tan20tan40 and tan60°=3=tan60°-tan60°tan20°tan40°+tan60°tan20°tan40°=tan60° =3

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Question 4:

If tan A=aa+1and tan B=12a+1, then the value of A + B is

(a) 0
(b) π2
(c) π3
(d) π4

Answer:

(d)  π4

tan(A+B) =tanA+tanB1-tanAtanB                  =aa+1+12a+11-aa+1(2a+1)                =  2a2+a+a+12a2+3a+1-a                =2a2+2a+12a2+2a+1                =1Therefore, A+B = tan-1(1) = π4.         

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Question 5:

If 3 sin x + 4 cos x = 5, then 4 sin x − 3 cos x =
(a) 0
(b) 5
(c) 1
(d) None of these

Answer:

(a) 0

3 sin x +4 cos x=535sin x+45cos x=1Let cos α=35 and  sin α=45. cos α sin x+sinαcos x=1 sin α+x=sinπ2α+x=π2x=π2-α        ....(1)We have to find the value of 4 sin x-3 cos x. 4 sinπ2-α -3 cosπ2-α            ...{From eq (1)} =4cos α-3sinα  =4 × 35-3×45                cos α=35 and  sin α=45 =0

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Question 6:

If in ∆ABC, tan A + tan B + tan C = 6, then cot A cot B cot C =

(a) 6
(b) 1
(c) 16
(d) None of these

Answer:

(c) 16
In triangle ABC,
A+B+C=πWe know that tanA+B+ C=tan A+tan B+tan C-tanA tanB tan C1-tan A tan B-tan B tanC-tan C tan Aand tan π=0. tan A+tan B+tan C-tanA tanB tan C=0tan A+tan B+tan C=tanA tanB tan C

If tan A+tan B+tan C =6,
tan A tan B tan C =6
1tanA tanB tanC=16cotA cotB cotC =16

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Question 7:

tan 3A − tan 2A − tan A =

(a) tan 3 A tan 2 A tan A
(b) −tan 3 A tan 2 A tan A
(c) tan A tan 2 A − tan 2 A tan 3 A − tan 3 A tan A
(d) None of these

Answer:

(a)  tan 3A tan 2A tan A

  3A=2A+Atan 3A=tan(2A+A)              =tan2A+tanA1-tan2AtanAtan 3A-tan3A tan2A tanA =tan 2A+tanAtan 3A-tan 2A- tanA=tan3A tan2A tanA

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Question 8:

If A + B + C = π, then tan A + tan B + tan Ctan A tan B tan Cis equal to
(a) tan A tan B tan C
(b) 0
(c) 1
(d) None of these

Answer:

(c) 1
π = 180°
Using tan(180 – A) = -tan A, we get:
C=π-(A+B)Now,tanA+tanB+tanCtanA tanB tanC=tanA+tanB+tanπ-(A+B)tanA tanB tanπ-(A+B)=tanA+tanB-tan(A+B)-tanA tanB tan(A+B)=tanA+tanB-tan A+tan B1-tanA tanB-tanA tanB×tan A+tan B1-tanA tanB 

=   tanA+tanB-tan2AtanB-tanA tan2B-tanA-tanB-tan2A tanB-tanA tan2B=-tan2AtanB-tanA tan2B-tan2AtanB-tanA tan2B=1



Page No 7.28:

Question 9:

If cos P=17and cos Q=1314, where P and Q both are acute angles. Then, the value of PQ is
(a) π6

(b) π3

(c) π4

(d) π12

Answer:

(b) 60⁰ = π3
cosP=17 , cosQ=1314
Therefore, sin P =1-149=437and  sinQ =1-169196=3314
Hence,  tan P =43,  tan Q =3313
cos(P-Q) =cosP cosQ + sinP sinQ
=17×1314+437×3314=13+3698
=4998
 cos P-Q=12 P-Q=cos-112 P-Q=60°

Hence, the correct answer is option B.

Page No 7.28:

Question 10:

If cot (α + β) = 0, sin (α + 2β) is equal to

(a) sin α
(b) cos 2 β
(c) cos α
(d) sin 2 α

Answer:

(a) sin α
   

Given:cot(α+β) =0cos(α+β)sin(α+β)=0cos(α+β)=0α+β=π2     
Therefore, sinα+2β=sin α+α+β=sin α

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Question 11:

cos 10° + sin 10°cos 10° - sin 10°=

(a) tan 55°
(b) cot 55°
(c) −tan 35°
(d) −cot 35°

Answer:

(a) tan 55°

cos10°+sin10°cos10°-sin10°=1+tan10°1-tan10°      Dividing the numerator and denominator by cos 10°=tan45°+tan10°1-tan45°×tan10°=tan(45°+10°)              Using tan(A+B) =tanA+tanB1-tanA tanB= tan55°

Page No 7.28:

Question 12:

The value of cos2 π6+x-sin2 π6-x is
(a) 12 cos 2 x
(b) 0
(c) -12 cos 2 x
(d) 12

Answer:

(a)  12cos 2x
cos2π6+x-sin2π6-x=cosπ6+x+π6-x cosπ6+x-π6+x   Using  cos(A+B) cos(A-B) =cos2A- sin2B=cos2π6cos2x=12cos2x          As cosπ3=12

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Question 13:

If tan θ1 tan θ2 = k, then cos θ1-θ2cos θ1+θ2=

(a) 1+k1-k
(b) 1-k1+k
(c) k+1k-1
(d) k-1k+1

Answer:

(a) 1+k1-k


cos(θ1-θ2)cos(θ1+θ2)=cosθ1cosθ2+sinθ1 sin θ2cosθ1cosθ2-sinθ1 sin θ2Dividing numerator and denominator by cos θ1cos θ2 , we get:
1+tanθ1tanθ21-tanθ1tanθ2=1+k1-k

Page No 7.28:

Question 14:

If sin (π cos x) = cos (π sin x), then sin 2 x =

(a) ±34

(b) ±43

(c) ±13

(d) none of these

Answer:

(c) ±13sin π cos x=cos π sin xAs  we know that sin x=-cos π2+x -cos π2+π cos x=cos π sin x-π2-π cos x=π sin xπ sin x-π cos x=π2 sin x- cos x=12Squaring both sides we get, sin2x+cos2x-2 sinx cosx=141-sin 2x=14sin 2x=13And we know that sin x=cos π2-xcos π2-π cos x=cos π sin xπ2-π cos x=π sin xπ sin x+π cos x=π2 sin x+cos x=12Squaring both sides we get, sin 2x+cos 2x+2sin x cos x=141+sin 2x=14sin 2x=13Therefore, sin 2x=±13

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Question 15:

If tanθ=12 and tanϕ=13, then the value of θ+ϕ is

(a) π6                               (b) π                               (c) 0                               (d) π4                              

Answer:

It is given that tanθ=12 and tanϕ=13.

Now,

tanθ+ϕ=tanθ+tanϕ1-tanθtanϕ                 =12+131-12×13                 =5656                 =1

θ+ϕ=π4                  tanπ4=1

Hence, the correct answer is option D.

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Question 16:

The value of cos (36° − A) cos (36° + A) + cos (54° + A) cos (54° − A) is

(a) sin 2A
(b) cos 2A
(c) cos 3A
(d) sin 3A

Answer:

(b)  cos 2A

cos(36°-A)cos(36°+A)+cos(54°+A)cos(54°-A)=cos(36°-A) cos(36°+A) +sin90°-(54°+A) sin90°-(54°-A)            Since sin(90°-θ)= cosθ=cos(36°-A)cos(36°+A)+sin(36°-A)sin(36°+A)=cos(36°+A-36°+A)          Using cos(A-B)  formula=cos 2A

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Question 17:

If tan (π/4 + x) + tan (π/4 − x) = a, then tan2 (π/4 + x) + tan2 (π/4 − x) =
(a) a2 + 1
(b) a2 + 2
(c) a2 − 2
(d) None of these

Answer:

(c) a2-2
Given:tanπ4+x+tanπ4-x =atanπ4+x+tanπ4-x2=a2tan2π4+x+tan2π4-x+2 tanπ4-x tanπ4+x =a2tan2π4+x+tan2π4-x=a2-2 tanπ4-x tanπ4+x  tan2π4+x+tan2π4-x=a2-2tan45°-tanx1+tan45° tanx×tan45°+tanx1-tan45° tanx tan2π4+x+tan2π4-x=a2-21°-tanx1+ tanx×1+tanx1- tanxtan2π4+x+tan2π4-x=a2-21-tan2x1-tan2xtan2π4+x+tan2π4-x=a2-2

Page No 7.28:

Question 18:

If tan (AB) = 1 and sec (A + B) = 23, the smallest positive value of B is

(a) 25 π24
(b) 19 π24
(c) 13π24
(d) 11 π24

Answer:

(b) 19 π24
Given:tan(A-B)=1 and sec(A+B) =23A-B=π4...(1)  and A+B=π6...(2)Adding these equations we get: 2A=π4+π6A=5π24Smallest possible value of B=π-5π24=19π24.

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Question 19:

If AB = π/4, then (1 + tan A) (1 − tan B) is equal to

(a) 2
(b) 1
(c) 0
(d) 3

Answer:

(a)  2
tan(A-B)=tanπ4tanA-tanB1+tanA tanB=1tanA-tanB=1+tanAtanB       ...(1) Now,(1+tanA)(1-tanB ) =1+tanA-tanB-tanA tanB                                     =1+1+tanAtanB-tanA tanB       Using eq (1)                                     = 2           

Page No 7.28:

Question 20:

The maximum value of sin22π3+x+sin22π3-x is
(a) 1/2
(b) 3/2
(c) 1/4
(d) 3/4

Answer:

(b) 32
2π3=120°
Let f(x) =sin2(90+30+x)+ sin2(90+30-x)            =cos(30+x)2+cos(30-x)2               Using sin(90+A) = cosA                =32cosx-12sinx2+32cosx+12sinx2                  = 34cos2x+14sin2x-32cosx sinx+ 34cos2x+14sin2x+32cosx sinx                  =32cos2x+12sin2x                  =321-sin2 x+12sin2x                  =32-32sin2 x+12sin2x                  =32-sin2xFor f(x) to be maximum, sin2 x must have minimum value, which is 0.  32 is the maximum value of fx.

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Question 21:

If cos (AB)=35and tan A tan B = 2, then

(a) cos A cos B=15
(b) cos A cos B=-15
(c) sin A sin B=-15
(d) sin A sin B=-15

Answer:

(a) 15
tanA tanB = sinA sinBcosA cosB=2    Given     ...(1) 

Also,cos(A-B)=35cosA cosB+sinA sinB=35 
 sin A sinB =35-cos Acos B     ...(2)  Substituting eq (2) in eq (1), we get: 
35-cosA cosBcosA cosB=2    

3cosA cosB=35

cosA cosB       =15   

Page No 7.28:

Question 22:

If tan 69° + tan 66° − tan 69° tan 66° = 2k, then k =

(a) −1
(b) 12
(c) -12
(d) None of these

Answer:

(c)-12

tan135° =tan(90°+45°)               = -tan45°               = -1Or, tan(69°+66°)=tan69°+tan66°1-tan69° tan66°-1 =tan69°+tan66°1-tan69° tan66°tan69°+tan66°-tan69°tan66°=-1Therefore,2k=-1k=-12



Page No 7.29:

Question 23:

If tanα=xx+1 and tanβ=12x+1, then α+β is equal to

(a) π2                               (b) π3                               (c) π6                              (d) π4                              

Answer:

It is given that tanα=xx+1 and tanβ=12x+1.

tanα+β=tanα+tanβ1-tanαtanβ                 =xx+1+12x+11-xx+1×12x+1                 =x2x+1+x+1x+12x+1x+12x+1-xx+12x+1                 =2x2+x+x+12x2+3x+1-x
                 =2x2+2x+12x2+2x+1=1

α+β=π4                  tanπ4=1

Hence, the correct answer is option D.



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