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  • 1 answers

Rishi Raj Singh Jadon 5 years, 11 months ago

10
  • 1 answers

Rolika Jain 6 years, 3 months ago

(Sin-1/sin)(cos-1/cos) =[(sin^2-1)/sin][(cos^2-1)cos] =(cos^2/sin)x(sin^2/cos) =sin x cos ......I hope it would be right one....
  • 2 answers
Reply me at 10 o'clock today
Ncert textbook and ncert exampler only.....ncert cover 90 percent....ncert exampler cover 10 percent......hope it help you bro....are you want to become a member of ⛩️Singh Empire......
  • 3 answers

Aayush Kumar 6 years, 3 months ago

All theorems are important.

Inayat Khan 6 years, 3 months ago

Only one que.

Isha Antil 6 years, 3 months ago

Pgt theorms
  • 1 answers

Varshini Palakurty 6 years, 3 months ago

1.5
  • 3 answers

Inayat Khan 6 years, 3 months ago

I will go with varshini.. It will have many solutions

Varshini Palakurty 6 years, 3 months ago

Infinitly many solutions

Ritik Madhwani 6 years, 3 months ago

1 solution
  • 1 answers

Varshini Palakurty 6 years, 3 months ago

Height=15 Base=15 Angle of depression =? Consider triangle abc b is interior opposite angle of depression Opposite =15m Adjacent =15 Tan=opp/adj Tan=15/15 Tan=1 Tan45=1 Angle of depression =45
  • 1 answers

Inayat Khan 6 years, 3 months ago

Putting values of m as 1;2;3... 4m'2-m=4(1)'2-2(1)=2. 4m'2-2m=4(2)'2-2(2)=12. 4m' 2-2m=4(3)'2-2(3)=30. Now;s1=2;s2=12 nd s3=30. Also s1=a1 ;s2-s1=a2. So a1=2 and a2=12-2=10. Therefore d=a2-a1=10-2=8. As ;an=a+(n-1)d. =107=2+(n-1)8. =107=2+8n-8. =112=8n. n=14. Hope you will further solve it and it will help you .
  • 1 answers

Gaurav Seth 6 years, 3 months ago

69, 67, 71

Let ( a - d ) , a , ( a + d ) are three parts are in A.p

according to the problem given ,

a - d + a + a + d = 207

3a = 207

a = 207 / 3

a = 69

product of two smaller parts = 4623

( a - d ) × a = 4623

( 69 - d ) × 69 = 4623

69 - d = 4623/69

69 - d = 1541/23

69 - d = 67

-d = 67 - 69

- d = - 2

d = 2

Therefore ,

required three parts are ,

a - d = 69 - 2 = 67

a = 69

a + d = 69 + 2 = 71

  • 1 answers

Gaurav Seth 6 years, 3 months ago

Qustion: If sin 3A = cos (A – 26°), where 3A is an acute angle, find the value of A.

Solution:

Sin 3A = cos (A-26 )

cos (90 - 3A) = cos (A-26) [ sinA = cos (90- A )
By comparison

90-3A = A-26

90+26=A+3A
116=4A

A= 29

  • 1 answers

Inayat Khan 6 years, 3 months ago

Can't understand wht u exactly wanted to know
  • 3 answers

Isha Antil 6 years, 3 months ago

Give an answer in short way

Deepti Singh 6 years, 3 months ago

Ncert book theorem 1.4
Prefer rs aggarwal .....
  • 2 answers

Inayat Khan 6 years, 3 months ago

Solve it by elimination.if u can't solve it i will help u
X=2 ;y=1
  • 1 answers

Asmilo Hall 6 years, 3 months ago

Given 1/(cosA+sinA-1)+1/(cosA+sinA+1)=cosecA+secA Read more on Sarthaks.com - https://www.sarthaks.com/31243/prove-that-1-cosa-sina-1-1-cosa-sina-1-coseca-seca
  • 0 answers
  • 1 answers

Ram ? 6 years, 3 months ago

10x^2+x-3
  • 1 answers

Ram ? 6 years, 3 months ago

2.1&2.2
  • 1 answers

Super Sai 6 years, 3 months ago

Which question
  • 3 answers

Shashi Kumar 6 years, 3 months ago

Learn basic formula of trigonometry chapter

Abhishek Yadav 6 years, 3 months ago

First Convert it into sin and cos

Krishna Agnihotri 6 years, 3 months ago

Just try to remember the basic concept of proving
  • 1 answers

Aks Julka 6 years, 3 months ago

X=-1

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