Ask questions which are clear, concise and easy to understand.
Ask QuestionPosted by Shalini Rajput 6 years, 11 months ago
- 1 answers
Posted by Shalini Rajput 5 years, 8 months ago
- 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
Posted by Jas K?? 5 years, 8 months ago
- 7 answers
Rohit Sahani??? 6 years, 11 months ago
Ashish Kumar 6 years, 11 months ago
Avinash Saigal 6 years, 11 months ago
Posted by Yogesh Parihar 6 years, 11 months ago
- 1 answers
Lavina Rajput ??? 6 years, 11 months ago
Posted by Durgesh Kumar 6 years, 11 months ago
- 1 answers
Posted by Khawa Khan 6 years, 11 months ago
- 1 answers
Posted by Megha Mishra 6 years, 11 months ago
- 0 answers
Posted by Harsh Chahal 6 years, 11 months ago
- 3 answers
Posted by Rajni Shah 6 years, 11 months ago
- 3 answers
Anshika & Aliens???? 6 years, 11 months ago
Posted by Avaneesh Sharma Mdjssv 6 years, 11 months ago
- 1 answers
Anshika & Aliens???? 6 years, 11 months ago
Posted by Digraj Chouhan 6 years, 4 months ago
- 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
Posted by Sailesh Gupta 6 years, 11 months ago
- 1 answers
Parneet Kaur Dhaliwal 6 years, 11 months ago
Posted by Avnish Pandey 6 years, 11 months ago
- 2 answers
Parneet Kaur Dhaliwal 6 years, 11 months ago
Posted by Digraj Chouhan 6 years, 6 months ago
- 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} .
Posted by Digraj Chouhan 6 years, 11 months ago
- 4 answers
Posted by Digraj Chouhan 6 years, 11 months ago
- 1 answers
Posted by Parneet Kaur Dhaliwal 6 years, 11 months ago
- 2 answers
Anjali Kumari? 6 years, 11 months ago
Posted by Parneet Kaur Dhaliwal 6 years, 11 months ago
- 3 answers
Jot Virk?♠ 6 years, 11 months ago
Anshika & Aliens???? 6 years, 11 months ago
Posted by Parneet Kaur Dhaliwal 6 years, 11 months ago
- 1 answers
Anshika & Aliens???? 6 years, 11 months ago
Posted by Jot Virk?♠ 6 years, 11 months ago
- 3 answers
Shivam Balodi 6 years, 11 months ago
Posted by Akash Maurya 6 years, 11 months ago
- 3 answers
Posted by Bahubali Gaming 6 years, 11 months ago
- 3 answers
Anshika & Aliens???? 6 years, 11 months ago
Avinash Saigal 6 years, 11 months ago
Posted by Anshika & Aliens???? 6 years, 11 months ago
- 2 answers
Parneet Kaur Dhaliwal 6 years, 11 months ago
Posted by Jannat Jahan 6 years, 11 months ago
- 0 answers
Posted by Payal Chatterjee 6 years, 6 months ago
- 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.
Posted by Mohammad Amaan 6 years, 11 months ago
- 1 answers
Avinash Saigal 6 years, 11 months ago
Posted by Sadhana Sadhana 6 years, 11 months ago
- 1 answers

myCBSEguide
Trusted by 1 Crore+ Students

Test Generator
Create papers online. It's FREE.

CUET Mock Tests
75,000+ questions to practice only on myCBSEguide app
myCBSEguide
Pihu Saini 6 years, 11 months ago
4Thank You