maths 1a

TS Inter Practice Papers 2021

TS Inter Maths 1A &1B Practice papers (Reduced Syllabus)

TS Inter Maths 1A &1B Practice papers (Reduced Syllabus)

 

TS Inter Maths 1A and 1B Practice papers as per reduced syllabus were designed by the ‘Basics in Maths‘ team.

These Practice papers to do help the intermediate First-year Maths students.

TS Inter Maths 1A and 1B Practice papers as per reduced syllabus are very useful in IPE examinations.

 


 

MATHS 1A PRACTICE PAPER – 1

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TS Inter Maths 1A and 1B Practice papers as per reduced syllabus were designed by the ‘Basics in Maths‘ team.

These Practice papers to do help the intermediate First-year Maths students.

TS Inter Maths 1A and 1B Practice papers as per   syllabus are very useful in IPE examinations.


 

 

MATHS 1B PRACTICE PAPER – 1

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PDF Files

PDF Files || Inter Maths 1A &1B || (New)

Inter Maths 1A &1B|| PDF Files (New)

 

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Inter Mathematics 1A ands 1B Pdf Files|  these   Files were designed by the ‘Basics in Maths‘ team. These notes to do help the intermediate First-year Maths students.

Inter Maths – 1A & 1B   PDF Files are very useful in IPE examinations.

These notes cover all the topics covered in the intermediate First-year Maths syllabus and include plenty of solutions to help you solve all the major types of Math problems asked in the

IPE examinations.  

 


MATHEMATICS 1A

 

Addition Of Vectors SAQ’S

Matrices

 

TS 10th class maths concept (E/M)

 

 

MATHEMATICS 1B 

DC’s and Dr’s

Tangents and Normals

Maxima and Minima

 

 

 

TS 6th Class Maths Concept


 

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1643133454126

Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S

Trigonometric Ratios(Qns.& Ans) V.S.A.Q.’S designed by the ‘Basics in Maths‘ team.These notes to do help intermediate First-year Maths students.

Inter Maths – 1A two marks questions and solutions are very useful in I.P.E examinations.


Trigonometric Ratios Up to Transformations

 Question 1

Find the value of sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10)

Sol:

 sin2(π/10) + sin2(4π/10) + sin2(6π/10) + sin2(9π/10)

= sin2(π/10) + sin2(π/2 – π/10) + sin2(π/2+ π/10) + sin2(π – π/10)

= sin2(π/10) + cos2(π/10) + cos2(π/10) + sin2(π/10)

= 1 + 1 = 2

 Question 2

If sin θ = 4/5 and θ not in the first quadrant, find the value of cos θ

Sol:

Given sin θ = 4/5 and θ not in the first quadrant

⇒ θ in the second quadrant

⇒ cos θ < 0

    cos2θ = 1 – sin2 θ

              =1 – (4/5)2

             = 1 – 16/25

∴cos θ   = – 3/5 (∵cos θ < 0)

 Question 3

If 3sin θ + 4 cos θ = 5, then find the value of 4 sin θ – 3cos θ

Sol:

Given, 3sin θ + 4 cos θ = 5

let 4 sin θ – 3cos θ = x

  (3sin θ + 4 cos θ )2 + (4 sin θ – 3cos θ)2 = 52 + x2

 9 sin2 θ + 16 cos2 θ + 12 sin θ cos θ + 16 sin2 θ + 9 cos2 θ – 12sin θ cis θ = 25 + x2

25 sin2 θ + 25 cos2 θ = 25 + x2

25 = 25 + x2

⇒ x2 = 0

 x = 0

∴ 4 sin θ – 3cos θ = 0

 Question 4

If sec θ + tan θ =Trigonometry up to Transformations 1, find the value of sin θ and determine the quadrant in which θ lies

Sol:

Given, sec θ + tan θ =  ———— (1)

 We know that sec2 θ – tan2 θ = 1

⇒ (sec θ + tan θ) (sec θ – tan θ) = 1

 sec θ – tan θ =Trigonometry up to Transformations 2

⇒ sec θ – tan θ = Trigonometry up to Transformations 3———— (2)

 (1) + (2)

⇒ (sec θ + tan θ) + (sec θ – tan θ) = Trigonometry up to Transformations 4

2sec θ =Trigonometry up to Transformations 5

 sec θ =Trigonometry up to Transformations 6

(1) – (2)

⇒ (sec θ + tan θ) – (sec θ – tan θ) = Trigonometry up to Transformations 7

2 tan θ = Trigonometry up to Transformations 8 ⇒ tan θ =Trigonometry up to Transformations 9

Now sin θ = tan θ ÷ sec θ =Trigonometry up to Transformatio

     Sin θ =Trigonometry up to Transformations 11

Since sec θ positive and tan θ is negative θ lies in the 4th quadrant.

 Question 5

Prove that cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16) = 1

Sol:

cot (π/16). cot (2π/16). cot (3π/16).… cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (4π/16). cot (5π/16) cot (6π/16) cot (7π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). cot (π/2 – 3π/16) cot (π/2 – 2π/16) cot (π/2 – π/16)

= cot (π/16). cot (2π/16). cot (3π/16). cot (π/4). tan (3π/16) tan (2π/16) tan (π/16)

= [cot (π/16). tan (π/16)] [cot (2π/16). tan (2π/16)] [cot (3π/16). tan (3π/16]. cot (π/4)

= 1.1.1.1

 =1

 

 Question 6

If cos θ + sin θ = Trigonometry up to Transformations 12cos θ, then prove that cos θ – sin θ =  sin θ

Sol:

Given, cos θ + sin θ = Trigonometry up to Transformations 12cos θ

Sin θ = Trigonometry up to Transformations 12 cos θ – cos θ

           = ( Trigonometry up to Transformations 12 – 1) cos θ

( Trigonometry up to Transformations 12 + 1) sin θ = ( Trigonometry up to Transformations 12 + 1) ( Trigonometry up to Transformations 12 – 1) cos θ

Trigonometry up to Transformations 12 sin θ + sin θ = cos θ

∴ cos θ – sin θ = Trigonometry up to Transformations 12 sin θ

 Question 7

Find the value of 2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ)

Sol:

2(sin6 θ + cos6 θ) – 3 (sin4 θ + cos4 θ)

= 2[(sin2 θ)3 + (cos2 θ)3] – 3[(sin2 θ)2 + (cos2)2

= 2[(sin2 θ + cos2 θ)3 – 3 sin2 θ cos2 θ (sin2 θ + cos2 θ)] – 3[(sin2 θ + cos2 θ)2 – 2 sin2 θ cos2 θ]

= 2[1 – 3 sin2 θ cos2 θ] – 3 [1 – 2 sin2 θ cos2 θ]

= 2 – 6 sin2 θ cos2 θ – 3 + 6 sin2 θ cos2 θ

= – 1

 Question 8

If tan 200 = λ, then show thatTrigonometry up to Transformations 13    

Sol:

Given tan 200 = λ

Trigonometry up to Transformations 19 =Trigonometry up to Transformations 14

                               =Trigonometry up to Transformations 15

                              =Trigonometry up to Transformations 16

                              =Trigonometry up to Transformations 18

 Question 9

If sin α + cosec α = 2, find the value of sinn α + cosecn α, n∈ Z

Sol:

Given sin α + cosec α = 2

 ⇒ sin α + 1/ sin α = 2

 ⇒Trigonometry up to Transformations 17  = 2

      sin2 α + 1= 2 sin α

       sin2 α – 2 sin α + 1= 0

  (sin α – 1 )2 = 0

⇒ sin α – 1 = 0

sin α = 1 ⇒ cosec α = 1

 sinn α + cosecn α = (1)n + (1)n =1 + 1 =2

∴ sinn α + cosecn α = 2

 Question 10

Evaluate sin2 Trigonometry up to Transformations 20+ cos2 Trigonometry up to Transformations 21  – tan2   Trigonometry up to Transformations 22

Sol:

 Trigonometry up to Transformations 23

Trigonometry up to Transformations 24

 

 Question 11

Find the value of sin 3300. cos 1200 + cos 2100. Sin 3000

Sol:

 sin 3300. cos 1200 + cos 2100. Sin 3000

=sin (3600 – 300). cos (1800 – 600) + cos (1800 + 300). sin (3600 – 600)

= (– sin 300). (– cos 600) + (– cos300). (– sin600)

= sin 300.  cos 600 + cos300.  Sin600

= sin (600 + 300) = sin 900

=1

 Question 12

Prove that cos4 α + 2 cos2 α Trigonometry up to Transformations 25= (1 – sin4 α)

Sol:

cos4 α + 2 cos2 α Trigonometry up to Transformations 25

= cos4 α + 2 cos2 α (1 – cos2 α)

= (cos2 α)2 + 2 (1 – sin2 α) (sin2 α)

= (1 – sin2 α)2 + 2 sin2 α – 2sin4 α

= 1 + sin4 α – 2 sin2 α + 2 sin2 α – 2sin4 α

= 1 – sin4 α

 Question 13

Eliminate θ from x = a cos3 θ and y = b sin3 θ

Sol:

Given x = a cos3 θ and y = b sin3 θ

 cos3 θ = x/a and sin3 θ = y/b

 cos θ = (x/a)1/3 and sin θ = (y/b)1/3

we know that sin2 θ + cos2 θ = 1

 ⇒ [(y/b)1/3]2 + [(x/a)1/3]2 = 1

   (x/a)2/3 + (y/b)2/3 = 1

 Question 14

Find the period of the following functions

Sol:

(i) f(x) = tan 5x

we know that period of tan kx =Trigonometry up to Transformations 26

⇒ period of tan 5x =Trigonometry up to Transformations 27

(ii) f(x) =Trigonometry up to Transformations 28

we know that period of Trigonometry up to Transformations 29= Trigonometry up to Transformations 31

period of Trigonometry up to Transformations 28 = Trigonometry up to Transformations 32

                                      =Trigonometry up to Transformations 33

(iii) f(x) = 2 sinTrigonometry up to Transformations 34 + 2 cosTrigonometry up to Transformations 35

period of sin Trigonometry up to Transformations 34 = Trigonometry up to Transformations 36 = 8

period of cos Trigonometry up to Transformations 35 = Trigonometry up to Transformations 37 = 6

period of given function is = LCM (8, 6) = 24

 

(iv) f(x) = tan (x + 4x + 9x +…. + n2x)

f(x) = tan (x + 4x + 9x +…. + n2x)

       = tan (1 + 4 + 9 + … + n2) x

= tanTrigonometry up to Transformations 38x

we know that period of tan kx =Trigonometry up to Transformations 26

Period of tan  =Trigonometry up to Transformations 39

                        = Trigonometry up to Transformations 40

 Question 15

Prove that sin2(52 ½)0 – sin2 (22 ½)0 =Trigonometry up to Transformations 41

Sol:

We know that sin2 A – sin2B = sin (A +B) sin (A – B)

 ⇒ sin2(52 ½)0 – sin2 (22 ½)0

= sin (52 ½+ 22 ½) sin (52 ½ – 22 ½)

 = sin 750 sin 300

 = Trigonometry up to Transformations 42      

∴ sin2(52 ½)0 – sin2 (22 ½)0 =Trigonometry up to Transformations 41

 

 Question 16

Prove that tan 700 – tan200 = 2 tan 500

Sol:

500 = 700 – 200

Tan 500 = tan (700 – 200)

We know that tan (A –B) =Trigonometry up to Transformations 43

  ⇒ Tan 500 =Trigonometry up to Transformations 44

 ⇒ tan 700 – tan 200 = tan 500 (1 + tan700 tan 200)

     tan 700 – tan 200 = tan 500 [1 + tan700 cot (900 – 200)]

     tan 700 – tan 200 = tan 500 [1 + tan700 cot 700]

     tan 700 – tan 200 = tan 500 [1 + 1]

∴ tan 700 – tan200 = 2 tan 500

 Question 17 

If sin α = Trigonometry up to Transformations 45, sin β =Trigonometry up to Transformations 46  and α, β are acute, show that α + β =Trigonometry up to Transformations 47

Sol:

Given sin α =    Trigonometry up to Transformations 45                                           sin β =Trigonometry up to Transformations 46

Trigonometry up to Transformations 48

 tan α = 1/3                                                          tan β = ½

tan (α + β) =Trigonometry up to Transformations 49

      Trigonometry up to Transformations 50 

  tan (α + β) = 1

∴ α + β =Trigonometry up to Transformations 47

 Question 18

Find tanTrigonometry up to Transformations 51 in terms of tan A

Sol:

 tan Trigonometry up to Transformations 51 =Trigonometry up to Transformations 52

                        =Trigonometry up to Transformations 53

 Question 19

Prove thatTrigonometry up to Transformations 54 = cot 360

Sol:

Trigonometry up to Transformations 54  = Trigonometry up to Transformations 55

(on dividing numerator and denominator by cos 90)

      = Trigonometry up to Transformations 56

   = tan (450 + 90)

    = tan 540

 = tan (900 – 360)

 = cot 360

 ∴Trigonometry up to Transformations 54  = cot 360      

 Question 20

Show that cos 420 + cos 780 + cos 1620 = 0

Sol:

cos 420 + cos 780 + cos 1620

= cos (600 – 180) + cos (600 + 180) + cos (1800 – 180)

=cos 600 cos180 + sin 600 sin 180 + cos 600 cos 180 – sin 600 sin 180 – cos 180

= 2 cos 600 cos 180 – cos 180

= 2 (1/2) cos 180 – cos 180

 = cos 180 – cos 180

= 0

 

 Question 21

Express Trigonometry up to Transformations 57sin θ + cos θ as a single of an angle

Sol:

Trigonometry up to Transformations 57sin θ + cos θ = 2(Trigonometry up to Transformations 58  sin θ + Trigonometry up to Transformations 59cos θ)

                                = 2(cos 300 sin θ + sin 300 cos θ)

                                = 2 sin (θ + 300)

 Question 22

Find the maximum and minimum value of the following functions

(i) 3 sin x –4 cos x

a= 3, b = –4 and c = 0

Trigonometry up to Transformations 60  

                                     = 5

∴ minimum value = –5 and maximum value = 5

(ii) cos (x + ) + 2  sin (x + ) – 3

a= 1, b = 2  and c = – 3

Trigonometry up to Transformations 61

 ∴ minimum value = –6 and maximum value = 0

 Question 23

Find the range of the function f(x) = 7 cos x – 24sin x + 5

Sol:

Given f(x) = 7 cos x – 24sin x + 5

a= 7, b = –24 and c = 5

Trigonometry up to Transformations 62

∴ Range = [–20, 30]    

 Question 24

Prove that sin2α + cos2 (α + β) + 2 sin α sin β cos (α + β) is independent of α

Sol:

sin2α + cos2 (α + β) + 2 sin α sin β cos (α + β)

= sin2α + cos (α + β) [ cos (α + β) +2 sin α sin β]

= sin2α + cos (α + β) [ cos α cos β – sin α sin β +2 sin α sin β]

=sin2α + cos (α + β) [ cos α cos β + sin α sin β]

=sin2α + cos (α + β) cos (α –β)

= sin2 α + cos2 α – sin2 β

=1 – sin2 β

= cos2 β

 Question 25

Simplify Trigonometry up to Transformations 63

Sol:

Trigonometry up to Transformations 63   =Trigonometry up to Transformations 64

                 =Trigonometry up to Transformations 65

                 = tan θ

 

Question 26

For what values of x in the first quadrantTrigonometry up to Transformations 66 is positive?

Sol:

Trigonometry up to Transformations 66 > 0 ⟹ tan 2x > 0

⟹ 0 < 2x < π/2 (∵ x is in first quadrant)

⟹ 0 < x < π/4

Question 27

If cos θ = Trigonometry up to Transformations 67 and π < θ < 3π/2, find the value of tan θ/2.

Sol:

cos θ = Trigonometry up to Transformations 67

π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4

tan θ/2 < 0

tan θ/2 =Trigonometry up to Transformations 68

               =– Trigonometry up to Transformations 69 (tan θ/2 < 0)

              =–Trigonometry up to Transformations 70

           = – 2

Question 28

If A is not an integral multiple of π/2, prove that cot A – tan A = 2 cot 2A.

Sol:

cot A – tan A = Trigonometry up to Transformations 71

                         =Trigonometry up to Transformations 72

                          =Trigonometry up to Transformations 73

                          =Trigonometry up to Transformations 74

                           =Trigonometry up to Transformations 75

                           = 2 cot 2A

Question 29

Evaluate 6 sin 200 – 8sin3 200

Sol:

6 sin 200 – 8sin3 200 = 2 (3 sin 200 – 4sin3 200)

                                       = 2 sin 3(200)

                              = 2 sin 600

                              = 2Trigonometry up to Transformations 76 

                              =Trigonometry up to Transformations 77

Question 30

Express cos6 A + sin6 A in terms of sin 2A.

Sol:

cos6 A + sin6 A

= (sin2 A)3 + (cos2 A)3

= (sin2 A + cos2 A)3 – 3 sin2 A cos2 A (sin2 A + cos2 A)

= 1 – 3 sin2 A cos2 A

=1 – ¾ (4 sin2 A cos2 A)

 = 1 – ¾ sin22 A

 

Question 31

If 0 < θ < π/8, show that Trigonometry up to Transformations 79  = 2 cos (θ/2)

Sol:

Trigonometry up to Transformations 79

Trigonometry up to Transformations 80

Trigonometry up to Transformations 81

 =2 cos (θ/2)

Question 32

Find the extreme values of cos 2x + cos2x

Sol:

cos 2x + cos2x = 2cos2 x– 1 + cos2 x

                              =3cos2 x – 1

We know that – 1 ≤ cos x ≤ 1

 ⟹ 0 ≤ cos2 x ≤ 1

      3×0 ≤ 3×cos2 x ≤ 3×1

      0– 1 ≤3 cos2 x – 1≤ 3– 1

   – 1≤3 cos2 x – 1≤ 2

     Minimum value = – 1

     Maximum value = 2

Question 33

Prove that Trigonometry up to Transformations 82 = 4

Sol:

Trigonometry up to Transformations 82

Trigonometry up to Transformations 83

= 4

Question 34

Prove that sin 780 + cos 1320 =Trigonometry up to Transformations 84

Sol:

 sin 780 + cos 1320 = sin 780 + cos (900 + 420)

                                      = sin 780 – sin 420

                                      = 2 cosTrigonometry up to Transformations 85  sinTrigonometry up to Transformations 86

                                     = 2 cos 600 sin 180

                                      = 2Trigonometry up to Transformations 87

                                      =Trigonometry up to Transformations 84

Question 35

Find the value of sin 340 + cos 640 – cos40

Sol:

sin 340 + cos 640 – cos40

= sin 340 –2 sinTrigonometry up to Transformations 88  sinTrigonometry up to Transformations 89

= sin 340 – 2sin 340 sin 300

= sin 340 – 2 sin 340 (1/2)

=sin 340 – sin 340

=0

 

Question 36

Prove that 4(cos 660 + sin 840) =Trigonometry up to Transformations 90

Sol:

4(cos 660 + sin 840)  

=4(cos 660 + sin (900 – 60)  

=4(cos 660 + cos (60)  

= 4[ 2 cosTrigonometry up to Transformations 91  cos Trigonometry up to Transformations 92]

=4[ 2 cosTrigonometry up to Transformations 93  cosTrigonometry up to Transformations 94 ]

=8 cos 360 cos 300

= 8Trigonometry up to Transformations 95

=Trigonometry up to Transformations 90

Question 35

Prove that (tan θ + cot θ)2 = sec2 θ + cosec2 θ = sec2 θ. cosec2 θ

Sol:

tan θ + cot θ =    Trigonometry up to Transformations 96

                         =Trigonometry up to Transformations 97

                         =Trigonometry up to Transformations 98

                         = sec θ. cosec θ

(tan θ + cot θ)2   = sec2 θ. cosec2 θ

sec2 θ + cosec2 θ =Trigonometry up to Transformations 99

                               Trigonometry up to Transformations 100

                              = sec2 θ. cosec2 θ


 

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