# solutions

## Sem – 2 Concept – Engineering Mathematics

Sem – 2 Concept

Education is the acquisition of knowledge, skill value, etc.
Education contributes to the development of society.
website has been given material for math Polytechnic Students.

Engineering Maths  Sem – 2 Concept solutions in PDF Files are designed by the ‘Basics In Maths” Team. These Pdf Files are very useful for students who are prepared for polytechnic examinations.

TS 6th Class Maths Concept

### Polytechnic-Engineering Mathematics -SEM-2-Concept.pdf

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TS 10th class maths concept (E/M)

## Sem -2 Solutions – Polytechnic Engineering Mathematics

Sem -2 Solutions

Education is the acquisition of knowledge, skill value, etc.
Education contributes to the development of society.
website has been given material for mathematics Polytechnic Students.

Polytechnic Engineering Mathematics – Sem -2 Solutions in PDF Files are designed by the ‘Basics In Maths” Team. These Pdf Files are very useful for students who are prepared for polytechnic examinations.

#### Tangents and Normals

Tangents and Normals solutions  PDF File is designed by the ‘Basics In Maths” Team. This Pdf File is very useful for students who are prepared for polytechnic examinations.

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### Rate Measure

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#### Maxima and Minima

The Maxima and Minim solutions  PDF File is designed by the ‘Basics In Maths” Team. This Pdf File is very useful for students who are prepared for polytechnic examinations.

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## 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 =, 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 =

⇒ sec θ – tan θ = ———— (2)

(1) + (2)

⇒ (sec θ + tan θ) + (sec θ – tan θ) =

2sec θ =

sec θ =

(1) – (2)

⇒ (sec θ + tan θ) – (sec θ – tan θ) =

2 tan θ =  ⇒ tan θ =

Now sin θ = tan θ ÷ sec θ =

Sin θ =

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 = cos θ, then prove that cos θ – sin θ =  sin θ

Sol:

Given, cos θ + sin θ = cos θ

Sin θ =  cos θ – cos θ

= (  – 1) cos θ

(  + 1) sin θ = (  + 1) (  – 1) cos θ

sin θ + sin θ = cos θ

∴ cos θ – sin θ =  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 that

Sol:

Given tan 200 = λ

=

=

=

=

###### 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

⇒  = 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 + cos2   – tan2

Sol:

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 = (1 – sin4 α)

Sol:

cos4 α + 2 cos2 α

= 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 =

⇒ period of tan 5x =

(ii) f(x) =

we know that period of =

period of =

=

(iii) f(x) = 2 sin + 2 cos

period of sin  =  = 8

period of cos  =  = 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

= tanx

we know that period of tan kx =

Period of tan  =

=

Question 15

Prove that sin2(52 ½)0 – sin2 (22 ½)0 =

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

=

∴ sin2(52 ½)0 – sin2 (22 ½)0 =

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) =

⇒ Tan 500 =

⇒ 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 = , sin β =  and α, β are acute, show that α + β =

Sol:

Given sin α =                                               sin β =

tan α = 1/3                                                          tan β = ½

tan (α + β) =

tan (α + β) = 1

∴ α + β =

#### Question 18

Find tan in terms of tan A

Sol:

tan  =

=

Question 19

Prove that = cot 360

Sol:

=

(on dividing numerator and denominator by cos 90)

=

= tan (450 + 90)

= tan 540

= tan (900 – 360)

= cot 360

∴  = 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 sin θ + cos θ as a single of an angle

Sol:

sin θ + cos θ = 2(  sin θ + cos 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

= 5

∴ minimum value = –5 and maximum value = 5

(ii) cos (x + ) + 2  sin (x + ) – 3

a= 1, b = 2  and c = – 3

∴ 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

∴ 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

Sol:

=

=

= tan θ

Question 26

For what values of x in the first quadrant is positive?

Sol:

> 0 ⟹ tan 2x > 0

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

⟹ 0 < x < π/4

Question 27

If cos θ = and π < θ < 3π/2, find the value of tan θ/2.

Sol:

cos θ =

π < θ < 3π/2 ⟹ π/2 < θ/2 < 3π/4

tan θ/2 < 0

tan θ/2 =

=–  (tan θ/2 < 0)

=–

= – 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 =

=

=

=

=

= 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

= 2

=

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  = 2 cos (θ/2)

Sol:

=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 = 4

Sol:

= 4

Question 34

Prove that sin 780 + cos 1320 =

Sol:

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

= sin 780 – sin 420

= 2 cos  sin

= 2 cos 600 sin 180

= 2

=

Question 35

Find the value of sin 340 + cos 640 – cos40

Sol:

sin 340 + cos 640 – cos40

= sin 340 –2 sin  sin

= 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) =

Sol:

4(cos 660 + sin 840)

=4(cos 660 + sin (900 – 60)

=4(cos 660 + cos (60)

= 4[ 2 cos  cos ]

=4[ 2 cos  cos ]

=8 cos 360 cos 300

= 8

=

Question 35

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

Sol:

tan θ + cot θ =

=

=

= sec θ. cosec θ

(tan θ + cot θ)2   = sec2 θ. cosec2 θ

sec2 θ + cosec2 θ =

= sec2 θ. cosec2 θ