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Vanga Goutham Reddy Class 12 Tuition trainer in Hyderabad

Vanga Goutham Reddy

experience 1 yrs of Exp
students 1 student
locationImg Begumpet, Hyderabad
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I am a engineering graduate student . I am good at all subjects. I am graduated from JNTUH. I want to teach the tuitions as I want to share my knowledge to other students.

Languages Spoken

Telugu Mother Tongue (Native)

English Proficient

Hindi Proficient

Education

Anurag group of institutions 2018

Bachelor of Technology (B.Tech.)

Address

Begumpet, Hyderabad, India - 500016

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Teaches

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

1

Board

State, CBSE, International Baccalaureate, ISC/ICSE

Subjects taught

Mathematics, Geography, Telugu, Electronics, History, Engineering Graphics, Computer Science, Biology, English, Sanskrit, Political Science, Chemistry, Physics, Economics

Taught in School or College

No

Class 10 Tuition
1 Student

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Physics, English, Mathematics, Science, Sanskrit, Biology, Telugu, Social Science, Geography, Chemistry, History and Civics, Economic Application, English Literature

Taught in School or College

No

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, CBSE, ISC/ICSE

Subjects taught

Physics, English, Economics, Biotechnology, Mathematics, Geography, History, Electronics, Biology, Computer Science, Telugu, Engineering Graphics, Chemistry

Taught in School or College

No

BTech Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in BTech Tuition

1

BTech Electrical & Electronics subjects

(Dc- Ac) Direct Current-Alternatiing Current System Interaction, Design Of Embedded Controllers, Embedded Processor Architecture, Power System Protection, Regulation & Security, Switchgear And Protection, Electrical Circuit Analysis, Electromagnetic Theory, Control Systems, Analog And Digital Communication, Data Structures & Algorithms, Direct Energy Conversion, Power System Engineering, Applications of Digital Signal Processing (DSP), High-Voltage Direct Current (Hvdc) Transmission, Electric Drives, Alternative Energy Sources, High-Voltage Engineering, Discrete Fourier Transforms And Digital Filter Design, Electric Energy Systems, Power Electronics, Circuit Theory

BTech Branch

BTech Electrical & Electronics, BTech 1st Year Engineering

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions

Taught in School or College

No

BTech 1st Year subjects

Engineering Graphics, Advanced Mathematics (M2), Basic Electrical Technology, Computer science, Engineering Physics, Environmental Studies, Engineering Mathematics (M1), Basic Electronics, Engineering Chemistry

Reviews

No Reviews yet!

Answers by Vanga Goutham Reddy

Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3

Ask a Question

Post a Lesson

Slope of normal and slope of tangent are perpendicular to each other.e.g., slope of normal x = 1 - a sinθdifferentiate x with respect to θ,dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)y = bcos²θdifferentiate y with respect to θ,dy/dθ = b. d(cos²θ)/dθ=... ...more

Slope of normal and slope of tangent are perpendicular to each other.
e.g., slope of normal 

x = 1 - a sinθ
differentiate x with respect to θ,
dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)
y = bcos²θ
differentiate y with respect to θ,
dy/dθ = b. d(cos²θ)/dθ
= b. 2cosθ. (-sinθ)
= -2bsinθ.cosθ --------(2)

dividing equations (2) by (1),

dividing equations (2) by (1),
dy/dx = -2bsinθ.cosθ/-acosθ = 2b/a sinθ
at θ = π/2 , dy/dx = 2b/a sinπ/2 = 2b/a
so, slope of normal = -1/slope of tangent
= -1/(2b/a) = -a/2b

Answers 5 Comments
Dislike Bookmark

Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Determinants/NCERT Solutions/Miscellaneous Exercise 4

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Post a Lesson

(x+a)× - x + x =0 (x+a) × - x + x = 0 (x+a) - ax² -ax² =0 2ax² + a²x + 2a²x + a³ -2ax² = 0 3a²x +a³ = 0 a² (3x + a) = 0 a != 0, ( 3x + a)= 0 3x = -a ...more

(x+a)×[(x+a)² - x²] - x [x(x+a) - x²] + x [ x²- x(x+a)]  =0

(x+a) ×[ x²+2ax+a² - x²] - x[ x² + ax -x²] + x [ x² - x² - ax] = 0

(x+a) [2ax + a²] - ax² -ax² =0

2ax² + a²x + 2a²x + a³ -2ax² = 0

3a²x +a³ = 0

a² (3x + a) = 0

a != 0, ( 3x + a)= 0

3x = -a

 

Answers 3 Comments
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Teaches

Class 12 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

1

Board

State, CBSE, International Baccalaureate, ISC/ICSE

Subjects taught

Mathematics, Geography, Telugu, Electronics, History, Engineering Graphics, Computer Science, Biology, English, Sanskrit, Political Science, Chemistry, Physics, Economics

Taught in School or College

No

Class 10 Tuition
1 Student

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, ICSE, CBSE

Subjects taught

Physics, English, Mathematics, Science, Sanskrit, Biology, Telugu, Social Science, Geography, Chemistry, History and Civics, Economic Application, English Literature

Taught in School or College

No

Class 11 Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Board

State, CBSE, ISC/ICSE

Subjects taught

Physics, English, Economics, Biotechnology, Mathematics, Geography, History, Electronics, Biology, Computer Science, Telugu, Engineering Graphics, Chemistry

Taught in School or College

No

BTech Tuition

Class Location

Online class via Zoom

Student's Home

Tutor's Home

Years of Experience in BTech Tuition

1

BTech Electrical & Electronics subjects

(Dc- Ac) Direct Current-Alternatiing Current System Interaction, Design Of Embedded Controllers, Embedded Processor Architecture, Power System Protection, Regulation & Security, Switchgear And Protection, Electrical Circuit Analysis, Electromagnetic Theory, Control Systems, Analog And Digital Communication, Data Structures & Algorithms, Direct Energy Conversion, Power System Engineering, Applications of Digital Signal Processing (DSP), High-Voltage Direct Current (Hvdc) Transmission, Electric Drives, Alternative Energy Sources, High-Voltage Engineering, Discrete Fourier Transforms And Digital Filter Design, Electric Energy Systems, Power Electronics, Circuit Theory

BTech Branch

BTech Electrical & Electronics, BTech 1st Year Engineering

Type of class

Regular Classes

Class strength catered to

One on one/ Private Tutions

Taught in School or College

No

BTech 1st Year subjects

Engineering Graphics, Advanced Mathematics (M2), Basic Electrical Technology, Computer science, Engineering Physics, Environmental Studies, Engineering Mathematics (M1), Basic Electronics, Engineering Chemistry

No Reviews yet!

Answers by Vanga Goutham Reddy

Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Application of Derivatives/NCERT Solutions/Exercise 6.3

Ask a Question

Post a Lesson

Slope of normal and slope of tangent are perpendicular to each other.e.g., slope of normal x = 1 - a sinθdifferentiate x with respect to θ,dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)y = bcos²θdifferentiate y with respect to θ,dy/dθ = b. d(cos²θ)/dθ=... ...more

Slope of normal and slope of tangent are perpendicular to each other.
e.g., slope of normal 

x = 1 - a sinθ
differentiate x with respect to θ,
dx/dθ = 0 - a.d(sinθ)/dθ = - a.cosθ ------(1)
y = bcos²θ
differentiate y with respect to θ,
dy/dθ = b. d(cos²θ)/dθ
= b. 2cosθ. (-sinθ)
= -2bsinθ.cosθ --------(2)

dividing equations (2) by (1),

dividing equations (2) by (1),
dy/dx = -2bsinθ.cosθ/-acosθ = 2b/a sinθ
at θ = π/2 , dy/dx = 2b/a sinπ/2 = 2b/a
so, slope of normal = -1/slope of tangent
= -1/(2b/a) = -a/2b

Answers 5 Comments
Dislike Bookmark

Answered on 05/09/2019 Learn CBSE - Class 12/Mathematics/Determinants/NCERT Solutions/Miscellaneous Exercise 4

Ask a Question

Post a Lesson

(x+a)× - x + x =0 (x+a) × - x + x = 0 (x+a) - ax² -ax² =0 2ax² + a²x + 2a²x + a³ -2ax² = 0 3a²x +a³ = 0 a² (3x + a) = 0 a != 0, ( 3x + a)= 0 3x = -a ...more

(x+a)×[(x+a)² - x²] - x [x(x+a) - x²] + x [ x²- x(x+a)]  =0

(x+a) ×[ x²+2ax+a² - x²] - x[ x² + ax -x²] + x [ x² - x² - ax] = 0

(x+a) [2ax + a²] - ax² -ax² =0

2ax² + a²x + 2a²x + a³ -2ax² = 0

3a²x +a³ = 0

a² (3x + a) = 0

a != 0, ( 3x + a)= 0

3x = -a

 

Answers 3 Comments
Dislike Bookmark
x

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