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    NCERT Solutions for Class 10 Maths – Introduction to Trigonometry

    Exercise 8.1


    Q1) In ABC , 90 at B, AB=24cm, BC = 7cm.

    Determine:

    (i)sin(A), cos(A)

    (ii) sin(C), cos(C)

    Ans.) In ABC , B=90

    By Applying Pythagoras theorem, we get

    AC2=AB2+BC2

    (24)2+72 =(576+49)

    AC2 = 625cm2

    à AC = 25cm

    (i) sin(A) = BC/AC = 7/25

    Cos(A) = AB/AC = 24/25

    (ii) sin(C) = AB/AC =24/25

    cos(C) = BC/AC = 7/25

     

    Q2) In the given figure find tan(P) – cot(R)

    Ans.) PR = 13cm,PQ = 12cm and QR = 5cm

    According to Pythagorean theorem,

    132=QR2+122 169=QR2+144 QR2=169144=25 QR=25−−√=5

    tan(P) = oppositesideadjacentside=QRPQ=512

    cot(P) = adjacentsideoppositeside = PQQR = 512

    tan(P) – cot(R) = 512512=0

    Therefore ,tan(P) – cot(R) = 0

     

    Q3) If sin(A) = 3/4, calculate cos(A) and tan(A)

    Ans.) Let ABC , be a right-angled triangle, right-angled at B.

    We know that sin(A) = BC/AC = 3/4

    Let BC be 3k and AC will be 4k where k is a positive real number.

    By Pythagoras theorem we get,

    AC2=AB2+BC2

     

    (4k)2=AB2+(3k)2

     

    16k29k2=AB2

     

    AB2=7k2

     

    AB=7–√k

     

    cos(A) = AB/AC = 7–√k/4k=7–√/4

    tan(A) = BC/AB =3k/7–√=3/7–√

     

    Q4) In question given below 15cot(A) = 8 ,find sin A and sec A.

    Ans.)  Let ABC be a right angled triangle, right-angled at B.

    We know that cot(A) = AB/BC = 8/15

    Given

    Let AB side be 8k and BC side 15k

    Where k is positive real number

    By Pythagoras theorem we get,

    AC2=AB2+BC2

     

    AC2=(8k)2+(15k)2

     

    AC2=64k2+225k2

     

    AC2=289k2

    AC = 17k

    sin(A) = BC/AC = 15k/17k = 15/7

    sec(A) =AC/AB =17k/8k = 17/8

     

    Q5) Given sec Ѳ =13/12, calculate all other trigonometric ratios.

    Ans.) Let  ABC be right-angled triangle, right-angled at B.

    We know that sec Ѳ =OP/OM =13/12(Given)

    Let side OP be 13k and side OM will be 12k where k is positive real number.

    By Pythagoras theorem we get,

    OP2=OM2+MP2

     

    (13k)2=(12k)2+MP2

     

    169(k)2144(k)2=MP2

     

    MP2=25k2

    MP = 5

    Now,

    sin Ѳ = MP/OP = 5k/13k =5/13

    cos Ѳ = OM/OP = 12k/13k = 12/13

    tan Ѳ = MP/OM = 5k/12k = 5/12

    cot Ѳ = OM/MP = 12k/5k = 12/5

    cosec Ѳ = OP/MP = 13k/5k = 13/5

     

    Q6) If A and B are acute angles such that

     cos(A) = cos(B), then show A =B .

    Ans.) Let  ABC in which CDAB .

    A/q,

    cos(A) = cos(B)

    à AD/AC = BD/BC

    à AD/BD = AC/BC

    Let  AD/BD =AC/BC =k

    AD =kBD …. (i)

    AC=kBC  …. (ii)

    By applying Pythagoras theorem in CAD and CBD we get,

    CD2=AC2AD2 ….(iv)

    From the equations (iii) and (iv) we get,

    AC2AD2=BC2BD2 AC2AD2=BC2BD2 k2(BC2BD2)=BC2BD2 k2=1

    Putting this value in equation (ii) , we obtain

    AC = BC

    A=B (Angles opposite to equal side are equal-isosceles triangle)

     

    Q7) If  cot Ѳ = 7/8, evaluate :

    (i) (1+sin Ѳ)(1-sin Ѳ) / (1+cos Ѳ)(1-cos Ѳ)

    (ii) cot2Θ

    Ans.) Let ABC in which  B=90

    and C=Θ

    A/q,

    cot Ѳ =BC/AB = 7/8

    Let BC = 7k and AB = 8k, where k is a positive real number

    According to Pythagoras theorem in ABC we get.

     

    AC2=AB2+BC2

     

    AC2=(8k)2+(7k)2

     

    AC2=64k2+49k2

     

    AC2=113k2

     

    AC=113−−−√k

     

    sin Ѳ = AB/AC = 8k/113−−−√k=8/113−−−√

    and cos Ѳ = BC/AC = 7k/113−−−√k=7/113−−−√

     

    (i) (1+sin Ѳ)(1-sinѲ)/(1+cosѲ)(1-cos Ѳ) = (1sin2Θ)/(1cos2Θ)

    = 1(8/113−−−√)2/1(7/113−−−√)2

    = {1-(64/113)}/{1-(49/113)} = {(113-64)/113}/{(113-49)/113} = 49/64

     

    (ii) cot2Θ=(7/8)2=49/64

     

    Q8) If 3cot(A) = 4/3, check whether (1tan2A)/(1+tan2A)=cos2Asin2A or not.

    Ans.) Let ABC in which B=90

    A/q,

    cot(A) = AB/BC = 4/3

    Let AB = 4k an BC =3k, where k is a positive real number.

    AC2=AB2+BC2

     

    AC2=(4k)2+(3k)2

     

    AC2=16k2+9k2

     

    AC2=25k2

     

    AC=5k

     

    tan(A) = BC/AB = 3/4

    sin(A) = BC/AC = 3/5

    cos(A) = AB/AC = 4/5

    L.H.S. = (1tan2A)(1+tan2A)=1(3/4)2/1+(3/4)2=(19/16)/(1+9/16)=(169)/(16+9)=7/25

    R.H.S. =cos2Asin2A=(4/5)2(3/4)2=(16/25)(9/25)=7/25

    R.H.S. =L.H.S.

    Hence, (1tan2A)/(1+tan2A)=cos2Asin2A

     

    Q9) In triangle EFG, right-angled at F, if tan E =1/√3 find the value of:
    (i) sin EcosG + cosE sin G
    (ii) cosEcosG – sin E sin G

    Answer

    LetΔEFG in which F=90, E/q

    tanE=FCEF tanE=FCEF=13

    Where k is the positive real number of the problem

    By Pythagoras theorem in ΔEFG we get:

    EG2=EF2+FG2 EG2=(3k−−√2))+K2 EG2=3k2+K2 EG2=4k2 EG=2K

     

    sinE = FG/EG = 1/2

    cosE = EF/EG =  32  ,
    sin G = EF/EG = 32 cosE = FG/EG = 1/2
    (i) sin EcosG + cosE sin G = (1/2\ast1/2) + (3232)= 1/4+3/4 = 4/4 = 1
    (ii) cosEcosG – sin E sin C = (3212)(3212)= (34)(34)= 0

     

    Q10)In Δ MNO, right-angled at N, MO + NO = 25 cm and MN = 5 cm. Determine the values of sin M, cos M and tan M.

    Answer

    Given that, MO + NO = 25 , MN = 5
    Let MO be x.  ∴ NO = 25 – x

    By Pythagoras theorem ,
    MO2=MN2+NO2
    X2=52+(25x)2
    50x = 650
    x = 13
    ∴ MO = 13 cm
    NO = (25 – 13) cm = 12 cm

    sinM = NO/MO = 12/13

    cosM = MN/MO = 5/13

    tanM = NO/MN = 12/5

     

    Q11)  State whether the following are true or false. Justify your answer.
    (i) The value of tan M is always less than 1.
    (ii) secM = 12/5 for some value of angle M.
    (iii) cosM is the abbreviation used for the cosecant of angle M.
    (iv) cot M is the product of cot and M.
    (v) sin θ = 4/3 for some angle θ.

    Answer

    (i) False.

    In ΔMNC in which N = 90,

    MN = 3, NC = 4 and MC = 5

    Value of tan M = 4/3 which is greater than.

    The triangle can be formed with sides equal to 3, 4 and hypotenuse = 5 as

    it will follow the Pythagoras theorem.

    MC2=MN2+NC2
    52=32+42
    25 = 9 + 16
    25 = 25

    (ii) True.
    Let a ΔMNC in which ∠N = 90º,MC be 12k and MB be 5k, where k is a positive real number.
    By Pythagoras theorem we get,
    MC2=MN2+NC2
    (12k)2=(5k)2+NC2
    NC2+25k2=144K2
    NC2=119k2

    Such a triangle is possible as it will follow the Pythagoras theorem.
    (iii) False.

    Abbreviation used for cosecant of angle M is cosec M.cosM is the abbreviation used for cosine of angle M.

    (iv) False.

    cotM is not the product of cot and M. It is the cotangent of M.
    (v) False.

    sinΘ = Height/Hypotenuse

    We know that in a right angled triangle, Hypotenuse is the longest side.

    ∴ sinΘwill always less than 1 and it can never be 4/3 for any value of Θ.

    Exercise 8.2

    1) Calculate the following:

    • sin60cos30+sin30cos60

     

    • 2tan245+co230sin260

     

    • cos45(sec30+cosec30)

     

    • (sin30+tan45cosec60)(sec30+cos60+cot45)

     

    • (5cos260+4sec230tan245)(sin230+cos230)

     

    Ans.- (i) sin60cos30+sin30cos60

    = (32×32)+(12×12)=34+14=44=1

     

    (ii) 2tan245+co230sin260

    =2×(1)2+(32)2(32)2=2

     

    (iii) cos45(sec30+cosec30)

    = 1223+2=12(2+23)3

    = 32×(2+23)=322+26

     

    = 3(2622)(26+22)(2622)

     

    = 23(62)(262 (22)2)

     

    23(62)248=23(62)16

     

    3(62)8=(186)8=(326)8

     

    (iv)  (sin 30° + tan 45° – cosec 60°)/(sec 30° + cos 60° + cot 45°)

    = (12+12323+12+1)

    = (322332+23)

    = (334)2(33)242

    = (27+16243)(2716)

    = (43243)11

     

    (v) (5cos260° + 4sec230° – tan245°)/(sin230° + cos230°)

    = 5(12)2+4(23)212(12)2+(32)2

    = (54+1631)(14+34)

    = (15+6412)1244

    =6712

     

    2) Find the correct answer and explain your choice:

     (i)  2tan301+tan230 =

              (A) sin 60 (B) cos 60 (C) tan 60 (D)        sin 30

     

     (ii) 1tan2451+tan230 =

    tan 90 (B) 1  (C) sin 45  (D) 0

     

    (iii) sin 2P = 2 sin P is true when P =

    0 (B)  30    (C)  45   (D)  60

     

    (iv)    2tan301tan230 =

    cos 60 (B)  sin 60   (C)  tan 60     (D)  sin 30   

     

    Ans.-

    (i)  (A) IS correct.

    2tan301+tan230 = 2(1)31+(13)2

    (23)1+13=(23)43 =643=32=sin60

     

    (ii)(D) is correct

    1tan2451+tan230

    = (112)(1+12)=02=0

     

    (iii) (A) is correct

    sin 2P = 2 sin P is true when

    P = sin 2P = sin 0° = 0
    2 sin P = 2sin 0° = 2×0 = 0

    or,

    sin 2P = 2sin PcosP

    =>2sin PcosP = 2 sin P

    =>2cos P = 2 =>cosP = 1

    =>P = 0°

     

    (iv) (C) is correct

    2tan301tan230=2(131(13)2)

     

    (23)113=2323=3–√=tan60

     

    3) If tan (P + Q) = 3–√ and tan ( P – Q) = 13;00<P+Q<=90;P>Q
    , calculate P and Q

                    Ans:-     tan (P + Q) = 3–√

    =>tan (P + Q) = tan 60°

    => (P + Q) =  60°     … (i)

    =>tan (P – Q) = 13

    =>tan (P – Q) = 30°

    => (P – Q) = 30°     … (ii)

    Adding (i) and (ii), we get

    P + Q + P – Q = 60° + 30°

    2P = 90°

    => P = 45°

    Putting the value of P in equation (i)

    45° + Q = 60°

    => Q = 60° – 45° = 15°

    Hence, P = 45° and Q = 15°

     

    4) Check whether the given statements are true or false, also give a reason for your answer:

    (i) sin (P + Q) = sin P + sin Q.

    (ii) The value of sin θ increases as θ increases.

    (iii) The value of cos θ increases as θ increases.

    (iv) sin θ = cos θ for all values of θ.

    (v) cotP is not defined for P = 0°.

    Ans:-

    (i) False

    Let P = 30° and Q = 60°, then
    sin (P + Q) = sin (30° + 60°) = sin 90° = 1 and,
    sin P + sin Q = sin 30° + sin 60°

    = 12+32=1+32

     

    (ii) True

    Sin 0° = 0

    Sin 30° = 12

    Sin 45° = 12

    Sin 60° = 32

    Sin 90° = 1

    Thus, the value of sinθ increases as θ increases

     

    (iii) False

    Cos 0° = 1

    Cos 30° = 32

    Cos 45° = 12

    Cos 60° = 12

    Cos 90° = 0

    Thus, the value of Cosθ decreases as θ increases.

    (iv) True

    cotP=cosPSinP cot0=cos0Sin0=10=notdefined

     

    Exercise 8.3

    1) Calculate:

                    (i) sin18cos72

                    (ii) tan26cot64

                    (iii) cos 48° – sin 42°

                    (iv) cosec 31° – sec 59°

    Ans:-

    (i) sin18cos72

    = sin(9018)cos72

    = cos72cos72=1

     

    (ii) tan26cot64

    = tan(9036)cot64

    cot64cot64=1

     

    (iii) cos 48° – sin42°

    = cos (90° – 42°) – sin 42°

    = sin 42° – sin 42° = 0

    (iv) cosec 31° – sec 59°

    = cosec (90° – 59°) – sec 59°
    = sec 59° – sec 59° = 0

     

    2) Show that :

     (i) tan 48° tan 23° tan 42° tan 67° = 1

    (ii) cos 38° cos 52° – sin 38° sin 52° = 0

    Ans:-

    (i)tan 48° tan 23° tan 42° tan 67°
    = tan (90° – 42°) tan (90° – 67°) tan 42° tan 67°
    = cot 42° cot 67° tan 42° tan 67°
    = (cot 42° tan 42°) (cot 67° tan 67°) = 1×1 = 1

    (ii) cos 38° cos 52° – sin 38° sin 52°
    = cos (90° – 52°) cos (90°-38°) – sin 38° sin 52°
    = sin 52° sin 38° – sin 38° sin 52° = 0

     

    3) We have 2P = cot ( P – 18 ° ), where 2P is an acute angle, calculate the value of P.

    Ans:-     According to question,
    tan 2P = cot (P- 18°)
    =>cot (90° – 2P) = cot (P -18°)
    Equating angles,
    =>90° – 2P = P- 18°

    =>108° = 3P
    => P = 36

     

    4) If tan P = cot Q, prove that P + Q = 90°.

     AnswerAccording to question,

    tanP = cot Q
    =>tan P = tan (90° – Q)
    =>P = 90° – Q
    =>P + Q = 90°

     

    5) If the value of sec 4P = cosec (P – 20°), in which 4P is an acute angle, find the value of P.

    Ans:-According to question

    sec 4P = cosec (P – 20°)

    => cosec (90° – 4P) = cosec (P – 20°)

    Equating angles,
    => 90° – 4P= P- 20°
    => 110° = 5P
    => P = 22°

     

    Q6) If X,Y and Z are interior angles of a triangle XYZ, then show that

        sin (Y+Z/2) = cos X2

    Answer

    In a triangle, sum of all the interior angles

    X + Y + Z = 180

    Y + Z = 180 – X

    Y+Z2 = (180X)2

    Y+Z2 = (90X2)

    sin (Y+Z2) = sin (90X2)

    sin (Y+Z2) = cosX2

     

    Q7) Express sin 67 + cos 75 in terms of trigonometric ratios of angles between 0 and 45.

    Answer

    sin 67 + cos 75

    = sin (9023) + cos (9015)
    = cos 23 + sin 15

     

    Excercise 8.4

     

    Q1) Express the trigonometric ratios sin A, sec A and tan A in terms of cot A.

    Answer

    cosec2Acot2A=1
    cosec2A = 1 + cot2A
    1sin2A = 1 + cot2A
    sin2A = 1/(1+cot2A)
    sin A= ±11+cot2A
    Now,
    sin2A=11+cot2A
    1cos2A=11+cot2A
    cos2A = 111+cot2A
    cos2A = (11+cot2A)(1+cot2A)
    1sec2A = (cot2A)(1+cotA)
    secA = (1+cotA)(cot2A)

     

    secA=±1+cot2AcotA

     

    also,
    tan A = sinAcosAand cot A = cosAsinA

    tan A = 1cotA

     

    Q2) Write all the other trigonometric ratios of A in terms of sec A.

    Answer

    We know that,
    sec A = 1cosA
    cos A = 1secA
    also,
    cos2A + sin2A = 1
     sin2A = 1 – cos2A
     sin2A = 1 – (1sec2A)
     sin2A = (sec2A1)sec2A

      sin A=±sec2A1secA

    also,
    sin A = 1cosecA
    cosec A = 1sinA

    cosec A=±secAsec2A1
    Now,
    sec2Atan2A = 1
    tan2A = sec2A + 1

    tan A=sec2A+1−−−−−−−−√
    also,
    tan A = 1cotA
    cot A = 1tanA

      cot A=±1sec2A+1

     

    Q3 Evaluate :


    (i) (sin263+sin227)(cos217+cos273)
    (ii)  sin25cos65++cos25sin65

     Answer

    (i) (sin263+sin227)(cos217+cos273)

     

    = [sin2(9027)+sin227][cos2(9073)+cos273]
    =(cos227+sin227)(sin227+cos273)
    = 11 =1          ( becausesin2A+cos2A=1)

    (ii) sin25cos65++cos25sin65
    =sin(9025)cos65+cos(9065)sin65

    =cos65cos65+sin65sin65

     

    = cos65+sin65=1

    4) Choose the correct option. Justify your choice.
    (i) 9 sec2A – 9 tan2A =
    (A) 1                 (B) 9              (C) 8                (D) 0
    (ii) (1 + tan Θ + sec Θ) (1 + cot Θ – cosec Θ)
    (A) 0                 (B) 1              (C) 2                (D) – 1
    (iii) (secA + tanA) (1 – sinA) =
    (A) secA           (B) sinA        (C) cosecA      (D) cosA

     

    (iv) 1+tan2A1+cot2A=

    (A) sec2A

    (B) -1

    (C) cot2A

    (D) tan2A

    Answer

    (i) (B) is correct.

    sec2A– 9 tan2A

    = 9 (sec2Atan2A                 )
    = 9×1 = 9             ( because  sec2Atan2A = 1)

     

    (ii) (C) is correct

    (1 + tan θ + sec θ) (1 + cot θ – cosec θ)

    = (1 + sin θ/cos θ + 1/cos θ) (1 + cos θ/sin θ – 1/sin θ)

    = (cosθ+sin θ+1)/cos θ × (sin θ+cos θ-1)/sin θ

    = (cosθ+sin θ)2-12/(cos θ sin θ)

    = (cos2θ + sin2θ + 2cos θ sin θ -1)/(cos θ sin θ)

    = (1+ 2cos θ sin θ -1)/(cos θ sin θ)

    = (2cos θ sin θ)/(cos θ sin θ) = 2

     

    (iii) (D) is correct.

    (secA + tanA) (1 – sinA)

    = (1/cos A + sin A/cos A) (1 – sinA)

    = (1+sin A/cos A) (1 – sinA)

    = (1 – sin2A)/cos A

    = cos2A/cos A = cos A

     

    (iv) (D) is correct.

    1+tan2A/1+cot2A

    = (1+1/cot2A)/1+cot2A

    = (cot2A+1/cot2A)×(1/1+cot2A)

    = 1/cot2A = tan2A

     

    Q5) Prove the following identities, where the angles involved are acute angles for which theexpressions are defined.

    (i) (cosec θ – cot θ)= (1-cos θ)/(1+cos θ)

    (ii) cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A

    (iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ

    [Hint : Write the expression in terms of sin θ and cos θ]

    (iv) (1 + sec A)/sec A = sin2A/(1-cos A)

    [Hint : Simplify LHS and RHS separately]

    (v) (cos A–sin A+1)/(cosA+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.

    (vi)1+sinA1sinA−−−−−√=secA+tanA

    (vii) (sin θ – 2sin3θ)/(2cos3θ-cos θ) = tan θ
    (viii) (sin A + cosec A)+ (cos A + sec A)2 = 7+tan2A+cot2A
    (ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
    [Hint : Simplify LHS and RHS separately]
    (x) (1+tan2A/1+cot2A) = (1-tan A/1-cot A)2 = tan2A

    Answer

    (i) (cosecΘcotΘ)2 = (1-cos θ)/(1+cos θ)
    L.H.S. =  (cosecΘcotΘ)2

    =(cosec2Θ+cot2Θ2cosecΘcotΘ)

    =(1sin2Θ+cos2Θsin2Θ2cosΘsin2Θ)

    = (1 + cos2Θ – 2cos θ)/(1 – cos2Θ)
    = (1cosΘ)2 /(1 – cosθ)(1+cos θ)
    = (1-cos θ)/(1+cos θ) = R.H.S.

     

    (ii)  cos A/(1+sin A) + (1+sin A)/cos A = 2 sec A
    L.H.S. = cos A/(1+sin A) + (1+sin A)/cos A
    = [cos2A +(1+sinA)2]/(1+sin A)cos A
    = (cos2A + sin2A + 1 + 2sin A)/(1+sin A)cos A
    = (1 + 1 + 2sin A)/(1+sin A)cos A
    = (2+ 2sin A)/(1+sin A)cos A
    = 2(1+sin A)/(1+sin A)cos A
    = 2/cos A = 2 sec A = R.H.S.

     

    (iii) tan θ/(1-cot θ) + cot θ/(1-tan θ) = 1 + sec θ cosec θ
    L.H.S. = tan θ/(1-cot θ) + cot θ/(1-tan θ)
    = [(sin θ/cos θ)/1-(cos θ/sin θ)] + [(cos θ/sin θ)/1-(sin θ/cos θ)]
    = [(sin θ/cos θ)/(sin θ-cos θ)/sin θ] + [(cos θ/sin θ)/(cosθ-sin θ)/cos θ]
    = sin2Θ /[cos θ(sin θ-cos θ)] + cos2Θ /[sin θ(cos θ-sin θ)]
    = sin2Θ /[cos θ(sin θ-cos θ)] – cos2Θ /[sin θ(sin θ-cos θ)]
    = 1/(sin θ-cos θ) [(sin2Θ /cos θ) – (cos2Θ /sin θ)]
    = 1/(sin θ-cos θ) × [(sin3Θcos3Θ)/sin θ cos θ]
    = [(sin θ-cos θ)(sin2Θ +cos2Θ +sin θ cos θ)]/[(sin θ-cos θ)sin θ cos θ]
    = (1 + sin θ cos θ)/sin θ cos θ)
    = 1/sin θ cos θ + 1
    = 1 + sec θ cosec θ = R.H.S.

     

    (iv)  (1 + sec A)/sec A = sin2Θ /(1-cos A)
    L.H.S. = (1 + sec A)/sec A
    = (1 + 1/cos A)/1/cos A
    = (cos A + 1)/cos A/1/cos A
    = cos A + 1
    R.H.S. = sin2Θ /(1-cos A)
    = (1 –cos2Θ)/(1-cos A)
    = (1-cos A)(1+cos A)/(1-cos A)
    = cos A + 1
    L.H.S. = R.H.S.

     

    (v) (cos A–sin A+1)/(cosA+sin A–1) = cosec A + cot A,using the identity cosec2A = 1+cot2A.
    L.H.S. = (cos A–sin A+1)/(cosA+sin A–1)
    Dividing Numerator and Denominator by sin A,
    = (cos A–sin A+1)/sin A/(cosA+sin A–1)/sin A
    = (cot A – 1 + cosec A)/(cot A+ 1 – cosec A)
    = (cot A – cosec2A + cot2A + cosec A)/(cot A+ 1 – cosec A) (using cosec2Acot2A = 1)
    = [(cot A + cosec A) – (cosec2A – cot2A)]/(cot A+ 1 – cosec A)
    = [(cot A + cosec A) – (cosec A + cot A)(cosec A – cot A)]/(1 – cosec A + cot A)
    =  (cot A + cosec A)(1 – cosec A + cot A)/(1 – cosec A + cot A)
    =  cot A + cosec A = R.H.S.

     

    (vi)1+sinA1sinA−−−−−√=secA+tanA

    Dividing Numerator and Denominator of L.H.S. by cos A,

    = 1cosA+sinAcosA1cosAsinAcosA

     

    = secA+tanAsecAtanA

     

    = secA+tanAsecAtanAXsecA+tanAsecA+tanA

     

    =(secA+tanA)2sec2Atan2A

     

    =secA+tanA1

    = sec A + tan A = R.H.S.

     

    (vii) (sin θ – 2sin3Θ)/(2cos3Θ -cos θ) = tan θ
    L.H.S. = (sin θ – 2sin3Θ)/(2cos3Θ – cos θ)
    = [sin θ(1 – 2sin2Θ)]/[cos θ(2cos2Θ – 1)]
    = sin θ[1 – 2(1-cos2Θ)]/[cosθ(2cos2Θ-1)]
    = [sin θ(2cos2Θ -1)]/[cos θ(2cos2Θ -1)]
    = tan θ = R.H.S.

     

    (viii) (sinA+cosecA)2 + (cosA+secA)2 = 7+tan2A +cot2A
    L.H.S. =  (sinA+cosecA)2 + (cosA+secA)2
                   = (sin2A + cosec2A + 2 sin A cosec A) + (tcos2A + sec2A + 2 cos A sec A)
    = (sin2A + cos2A) + 2 sin A(1/sin A) + 2 cos A(1/cos A) + 1 + tan2A + 1 + cot2A
    = 1 + 2 + 2 + 2 + tan2A + cot2A
    = 7+tan2A+cot2A = R.H.S.

     

    (ix) (cosec A – sin A)(sec A – cos A) = 1/(tan A+cotA)
    L.H.S. = (cosec A – sin A)(sec A – cos A)
    = (1/sin A – sin A)(1/cos A – cos A)
    = [(1-sin2A)/sin A][(1-cos2A)/cos A]
    = (cos2A/sin A)×(sin2A/cos A)
    = cos A sin A
    R.H.S. = 1/(tan A+cotA)
    = 1/(sin A/cos A +cos A/sin A)
    = 1/[(sin2A+cos2A)/sin A cos A]
    = cos A sin A
    L.H.S. = R.H.S.

     

    (x)  (1+tan2A/1+cot2A) = (1tanA1cotA)2 =tan2A
    L.H.S. = (1+tan2A/1+cot2A)
    = (1+tan2A/1+1/tan2A)
    = 1+tan2A/[(1+tan2A)/tan2A]
    = tan2A

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    2017-09-28T13:20:44+00:00 Categories: CBSE|0 Comments
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