Yasaswi Nath

Baner, Pune, India - 411045

Yasaswi Nath photo

Yasaswi Nath

Tutor

Baner, Pune, India- 411045.

3.9
UrbanPro Rating

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Overview

I'have completed B.Tech from IIT(BHU) the previous year, so I'm pretty strong in Maths and Science. I've 3 years of experience.

Languages Spoken

Oriya

Hindi

English

Education

IIT (Banaras Hindu University) 2017

Bachelor of Technology (B.Tech.)

Address

Baner, Pune, India - 411045

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Demo Class

No

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Class 6 Tuition Overview

Class 6 Tuition

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 6 Tuition

1

Board

State, ICSE

ICSE Subjects taught

Chemistry, Physics, Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics, Science

Reviews

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Answers by Yasaswi Nath (2)

Answered on 07/02/2018 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

m = tanX + sinX n = tanX - sinX m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX m2-n2 = 4tanX.sinX now, mn = (tanX + sinX).(tanX - sinX) mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X) = sin2X. (sin2X/cos2X)... ...more

m = tanX + sinX

n = tanX - sinX

m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX

m2-n2 = 4tanX.sinX

now, mn = (tanX + sinX).(tanX - sinX)

mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X)

      =  sin2X. (sin2X/cos2X) =  sin2X.tan2X

(mn)1/2 = sinX.tanX

We know that, m2-n2 = 4tanX.sinX

now replacing "tanX.sinX",

m2-n= 4(mn)1/2

Answers 3 Comments
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Answered on 07/02/2018 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

m = tanX + sinX n = tanX - sinX m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX m2-n2 = 4tanX.sinX now mn = (tanX + sinX).(tanX - sinX) mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X) = sin2X. (sin2X/ ...more

m = tanX + sinX

n = tanX - sinX

m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX

m2-n2 = 4tanX.sinX

now mn = (tanX + sinX).(tanX - sinX)

mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X)

=  sin2X. (sin2X/

Answers 3 Comments
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Yasaswi NathDirections

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Class 6 Tuition 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 6 Tuition

1

Board

State, ICSE

ICSE Subjects taught

Chemistry, Physics, Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics, Science

Class 7 Tuition 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 7 Tuition

1

Board

State, ICSE

ICSE Subjects taught

Chemistry, Physics, Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics, Science

Class 8 Tuition 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 8 Tuition

1

Board

State, ICSE

ICSE Subjects taught

Chemistry, Physics, Mathematics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics, Science

Class 9 Tuition 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 9 Tuition

1

Board

ICSE, CBSE, State

CBSE Subjects taught

Science, Mathematics

ICSE Subjects taught

Chemistry, Mathematics, Physics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics, Science

Class 10 Tuition 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class 10 Tuition

1

Board

ICSE, CBSE, State

CBSE Subjects taught

Science, Mathematics

ICSE Subjects taught

Chemistry, Mathematics, Physics

Taught in School or College

Yes

State Syllabus Subjects taught

Mathematics, Science

Class I-V Tuition 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Class I-V Tuition

1

Fees

₹ 200 per hour

Board

ICSE, CBSE

CBSE Subjects taught

Science, Mathematics

ICSE Subjects taught

Science, Mathematics

Taught in School or College

No

Math Olympiad classes 3.9

Class Location

Student's Home

Tutor's Home

Online (video chat via skype, google hangout etc)

Years of Experience in Math Olympiad classes

1

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No Reviews yet! Be the first one to Review

Answers by Yasaswi Nath (2)

Answered on 07/02/2018 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

m = tanX + sinX n = tanX - sinX m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX m2-n2 = 4tanX.sinX now, mn = (tanX + sinX).(tanX - sinX) mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X) = sin2X. (sin2X/cos2X)... ...more

m = tanX + sinX

n = tanX - sinX

m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX

m2-n2 = 4tanX.sinX

now, mn = (tanX + sinX).(tanX - sinX)

mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X)

      =  sin2X. (sin2X/cos2X) =  sin2X.tan2X

(mn)1/2 = sinX.tanX

We know that, m2-n2 = 4tanX.sinX

now replacing "tanX.sinX",

m2-n= 4(mn)1/2

Answers 3 Comments
Dislike Bookmark

Answered on 07/02/2018 CBSE/Class 10/Mathematics Tuition/Class IX-X Tuition

m = tanX + sinX n = tanX - sinX m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX m2-n2 = 4tanX.sinX now mn = (tanX + sinX).(tanX - sinX) mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X) = sin2X. (sin2X/ ...more

m = tanX + sinX

n = tanX - sinX

m2-n2 = (tanX + sinX)2 - (tanX - sinX)2 = tan2X + sin2X + 2tanX.sinX - tan2X - sin2X +2tanX.sinX

m2-n2 = 4tanX.sinX

now mn = (tanX + sinX).(tanX - sinX)

mn = tan2X - sin2X = (sin2X/cos2X) - sin2X = sin2X((1/cos2X) -1) = sin2X((1-cos2X)/cos2X)

=  sin2X. (sin2X/

Answers 3 Comments
Dislike Bookmark

Yasaswi Nath describes himself as Tutor. He conducts classes in Class 10 Tuition, Class 6 Tuition and Class 7 Tuition. Yasaswi is located in Baner, Pune. Yasaswi takes at students Home. He has 1 years of teaching experience . Yasaswi has completed Bachelor of Technology (B.Tech.) from IIT (Banaras Hindu University) in 2017. HeĀ is well versed in Oriya, Hindi and English.

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