No products in the cart.

Ask questions which are clear, concise and easy to understand.

Ask Question
  • 1 answers

Pihu Saini 6 years, 11 months ago

Given, tan 20 * tan 40 * tan 80 = tan 40 * tan 80 * tan 20 = [{sin 40 * sin 80}/{cos 40 * cos 80}] *(sin 20/cos 20) = [{2 * sin 40 * sin 80}/{2 * cos 40 * cos 80}] *(sin 20/cos 20) = [{cos 40 - cos 120}/{cos 120 + cos 40}] *(sin 20/cos 20) = [{cos 40 - cos (90 + 30)}/{cos (90 + 30) + cos 40}] *(sin 20/cos 20) = [{cos 40 + sin30}/{-sin30 + cos 40}] *(sin 20/cos 20) = [{(2 * cos 40 + 1)/2}/{(-1 + cos 40)/2}] *(sin 20/cos 20) = [{2 * cos 40 + 1}/{-1 + cos 40}] *(sin 20/cos 20) = [{2 * cos 40 * sin 20 + sin 20}/{-cos 20 + cos 40 * cos 20}] = (sin 60 - sin 20 + sin 20)/(-cos 20 + cos 60 + cos 20) = sin 60/cos 60 = tan 60...
  • 1 answers

Sia ? 6 years, 4 months ago

LHS = tan 20° tan 40° tan 80° {tex}=\frac{\sin 20^{\circ} \sin 40^{\circ} \sin 80^{\circ}}{\cos 20^{\circ} \cos 40^{\circ} \cos 80^{\circ}}{/tex}
{tex}=\frac{\left(2 \sin 20^{\circ} \sin 40^{\circ}\right) \sin 80^{\circ}}{\left(2 \cos 20^{\circ} \cos 40^{\circ}\right) \cos 80^{\circ}}{/tex} 
{tex}=\frac{\left(\cos 20^{\circ}-\cos 60^{\circ}\right) \sin 80^{\circ}}{\left(\cos 60^{\circ}+\cos 20^{\circ}\right) \cos 80^{\circ}}{/tex} [{tex}\because{/tex} 2 sin a sin b = cos (a-b) - cos (a+b), 2 cos a cos b = cos (a+b) + cos (a-b)]
{tex}=\frac{\sin 80^{\circ} \cos 20^{\circ}-(1 / 2) \sin 80^{\circ}}{(1 / 2) \cos 80^{\circ}+\cos 80^{\circ} \cos 20^{\circ}}{/tex} [ {tex}\because \cos\;60^o=\frac12{/tex}]
{tex}=\frac{2 \sin 80^{\circ} \cos 20^{\circ}-\sin 80^{\circ}}{\cos 80^{\circ}+2 \cos 80^{\circ} \cos 20^{\circ}}{/tex}
{tex}=\frac{\sin 100^{\circ}+\sin 60^{\circ}-\sin 80^{\circ}}{\cos 80^{\circ}+\cos 100^{\circ}+\cos 60^{\circ}}{/tex}[{tex}\because{/tex} 2 sin a cos b = sin (a+b) + sin (a-b)]
{tex}=\frac{\sin \left(180^{\circ}-80^{\circ}\right)+\sin 60^{\circ}-\sin 80^{\circ}}{\cos 80^{\circ}+\cos \left(180^{\circ}-80^{\circ}\right)+\cos 60^{\circ}}{/tex}
{tex}=\frac{\sin 80^{\circ}+\sin 60^{\circ}-\sin 80^{\circ}}{\cos 80^{\circ}-\cos 80^{\circ}+\cos 60^{\circ}}{/tex} {tex}=\frac{\sin 60^{\circ}}{\cos 60^{\circ}}{/tex} = tan 60° = RHS

  • 7 answers

Rohit Sahani??? 6 years, 11 months ago

No bro u will fail if u get 15 marks in maths. Passing marks=27out of 80.passing %=33.

Ashish Kumar 6 years, 11 months ago

Out of 100 marks 20 will be given by school on oral,PT, etc. To pass only 33%out of 100 you will have to score. Be patient!!

Avinash Saigal 6 years, 11 months ago

If I suggest then Just be positive .Don't think negative that you will get less marks in board exams.????

Jas K?‍? 6 years, 11 months ago

Practical means

Ashish Kumar 6 years, 11 months ago

You may if you will get 15+ marks in Practical!!

Jas K?‍? 6 years, 11 months ago

But I am too weak in maths

Janice Dias 6 years, 11 months ago

No you need minimum 30 /80
  • 1 answers

Lavina Rajput ??? 6 years, 11 months ago

No. of white balls=15 let the no. of black balls = x probability of drawing a white ball = 15/(15+x) probability of drawing a black ball = x/(15+x) According to given condition- x/(15+x)=3*15/(15+x) x=3*15 =45 therefore the no. of black balls = 45
  • 1 answers

Durgesh Kumar 5 years, 8 months ago

Anewer
  • 0 answers
  • 3 answers

Augustya Singh 6 years, 11 months ago

n/2(2a+(n-1)d)

Shreyanka Das 6 years, 11 months ago

Another formula for Sn is n/2(a+an).............

D.J Alok 6 years, 11 months ago

Sn= n/2{2a+(n-1)d}
  • 1 answers

Sia ? 6 years, 4 months ago

Since the points are collinear, then,
 Area of triangle = 0
{tex}\frac { 1 } { 2 } \left[ x _ { 1 } \left( y _ { 2 } - y _ { 3 } \right) + x _ { 2 } \left( y _ { 3 } - y _ { 1 } \right) + x _ { 3 } \left( y _ { 1 } - y _ { 2 } \right) \right] = 0{/tex}
{tex}\frac { 1 } { 2 } [ x ( - 4 + 5 ) + ( - 3 ) ( - 5 - 2 ) + 7 ( 2 + 4 ) ] = 0{/tex}
x + 21 + 42 = 0
x = -63

  • 1 answers

Parneet Kaur Dhaliwal 6 years, 11 months ago

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio
  • 2 answers

D.J Alok 6 years, 11 months ago

Proved in ncert.....

Parneet Kaur Dhaliwal 6 years, 11 months ago

In a right ?, the sq. of hypotenuse is equal to the sum of the sq. of the other two sides
  • 1 answers

Sia ? 6 years, 6 months ago

(3, a) lies on 2x - 3y - 5 = 0
So, 2 {tex}\times{/tex} 3 - 3y - 5 = 0
1 = 3a
a = {tex}\frac 13{/tex}
2x - 3y - 5 = 0 cuts the x-axis at (x, 0).
{tex}\therefore{/tex} 2x -5 = 0 or, x = {tex}\frac 52{/tex}
or, Point is {tex}\left( \frac { 5 } { 2 } , 0 \right){/tex} .

  • 4 answers

Parneet Kaur Dhaliwal 6 years, 11 months ago

X=9/2 and Y=0 therefore (9/2,0)

Digraj Chouhan 6 years, 11 months ago

Chapter 7 coordinate geometry

Parneet Kaur Dhaliwal 6 years, 11 months ago

Sorry wrong ans

Parneet Kaur Dhaliwal 6 years, 11 months ago

12 units
  • 2 answers

Anjali Kumari? 6 years, 11 months ago

First of all u have to find ( Perpedicular) by pythagoras theorem. Value of P is root21. Then 4 + 4×(root21/2)^2 4+ 4×( 21/4) 4+21 25 Ans

Anjali Kumari? 6 years, 11 months ago

25
  • 3 answers

Jot Virk?♠ 6 years, 11 months ago

Statement ko point to point and word to word padho aur according to given points steps to step figure banate raho and figure bilkul correct banagi

Parneet Kaur Dhaliwal 6 years, 11 months ago

Thanks

Anshika & Aliens???? 6 years, 11 months ago

Learn from CBSE guide may it help u i had learn all difficult chapters from this site this is really helpful ??
  • 1 answers

Anshika & Aliens???? 6 years, 11 months ago

1.because 1/cosec=sin.In trigonometry ratios table the maximum value for sin is 1
  • 3 answers

Jot Virk?♠ 6 years, 11 months ago

By which identity

Shivam Balodi 6 years, 11 months ago

First divide each term by SinA and then solve it by trignometric identity

Jot Virk?♠ 6 years, 11 months ago

Plz solve this problem fastly
  • 3 answers

Shivam Balodi 6 years, 11 months ago

2√3

Anshika & Aliens???? 6 years, 11 months ago

If u will simplify than answer will be 2✓3

Anshika & Aliens???? 6 years, 11 months ago

H=6✓3/3
  • 3 answers

Anshika & Aliens???? 6 years, 11 months ago

7th term = (a+6d)                         11th term = (a+10d) (a+6d)7  since 7th term of the ap is 7 times               and the same for 11th term (a+6d)7=(A+10d)11                   (given) 7a + 42d =11a +110d                       (on solving) 7a - 11a = 110d - 42d -4a = -68d a= -17d we know 18th term = (a+17d)................................. (i) now we substitute the value of a in (i)  which emplies,                        -17d + 17d                      = 0 hence prove..

Sumit Rajput 6 years, 11 months ago

What is a sun light

Avinash Saigal 6 years, 11 months ago

7th term = (a+6d)                         11th term = (a+10d) (a+6d)7  since 7th term of the ap is 7 times               and the same for 11th term (a+6d)7=(A+10d)11                   (given) 7a + 42d =11a +110d                       (on solving) 7a - 11a = 110d - 42d -4a = -68d a= -17d we know 18th term = (a+17d)................................. (i) now we substitute the value of a in (i)  which emplies,                        -17d + 17d                      = 0 Thank you??
  • 2 answers

Parneet Kaur Dhaliwal 6 years, 11 months ago

So simple

Parneet Kaur Dhaliwal 6 years, 11 months ago

Yar my cbse guides me na math sub. Khol osme revision notes pade he os se padh le
  • 1 answers

Sia ? 6 years, 6 months ago

Consider equations (p - 3)x + 3y = p
and px + py = 12
For infinitely many solutions,
{tex}\frac{p - 3}{p}{/tex} = {tex}\frac{3}{p}{/tex} = {tex}\frac{p}{12}{/tex}...........(i)
Consider, {tex}\frac{3}{p}{/tex} = {tex}\frac{p}{12}{/tex} {tex}\Rightarrow{/tex} p2 = 36  {tex}\Rightarrow{/tex} p = {tex}\pm{/tex}6
For p = 6, from (i) {tex}\frac{3}{6}{/tex} = {tex}\frac{3}{6}{/tex} = {tex}\frac{6}{12}{/tex}, true
For p = -6, from (i) {tex}\frac{-9}{-6}{/tex} = {tex}\frac{3}{-6}{/tex} = {tex}\frac{-6}{12}{/tex}, false.
Hence, for p = 6, pair of linear equations has infinitely many solutions.

  • 1 answers

Avinash Saigal 6 years, 11 months ago

Let the distance covered by the peacock be AC = x m Hence the distance covered by the snake is DC = x m In right ΔABC, by Pythagoras theorem we have AC2 = AB2 + BC2 x2 = 92 + (27 – x)2 ⇒ x2 = 81 + 729 – 54x + x2 ⇒ 54x = 810 ∴ x = 15....... Thank you???

myCBSEguide App

myCBSEguide

Trusted by 1 Crore+ Students

Test Generator

Test Generator

Create papers online. It's FREE.

CUET Mock Tests

CUET Mock Tests

75,000+ questions to practice only on myCBSEguide app

Download myCBSEguide App