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You can check the pattern here : <a href="https://mycbseguide.com/cbse-syllabus.html">https://mycbseguide.com/cbse-syllabus.html</a>
Posted by Bishnu Prasad Patra Patra 6 years, 6 months ago
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Sia ? 6 years, 6 months ago
Given,
tan A = n tan B
{tex} \Rightarrow{/tex} tanB = {tex}\frac{1}{n}{/tex}tan A
{tex}\Rightarrow{/tex} cotB = {tex}\frac { n } { \tan A }{/tex}..........(1)
Also given,
sin A = m sin B
{tex}\Rightarrow{/tex} sin B = {tex}\frac{1}{m}{/tex}sin A
{tex}\Rightarrow{/tex} cosec B = {tex}\frac { m } { \sin A }{/tex}.....(2)
We know that, cosec2B - cot2B = 1, hence from (1) & (2) :-
{tex} \quad \frac { m ^ { 2 } } { \sin ^ { 2 } A } - \frac { n ^ { 2 } } { \tan ^ { 2 } A } = 1{/tex}
{tex}\Rightarrow \quad \frac { m ^ { 2 } } { \sin ^ { 2 } A } - \frac { n ^ { 2 } \cos ^ { 2 } A } { \sin ^ { 2 } A } = 1{/tex}
{tex}\Rightarrow \quad \frac { m ^ { 2 } - n ^ { 2 } \cos ^ { 2 } A } { \sin ^ { 2 } A } = 1{/tex}
{tex}\Rightarrow{/tex} m2 - n2cos2A = sin2A
{tex}\Rightarrow{/tex} m2 - n2cos2A = 1 - cos2A
{tex}\Rightarrow{/tex} m2 - 1 = n2cos2A - cos2A
{tex}\Rightarrow{/tex} m2 - 1 = (n2 - 1) cos2A
{tex}\Rightarrow \quad \frac { m ^ { 2 } - 1 } { n ^ { 2 } - 1 } ={/tex} cos2A
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Sia ? 6 years, 6 months ago
Using distance formula, we obtain
SP = {tex}\sqrt { \left( a t ^ { 2 } - a \right) ^ { 2 } + ( 2 a t - 0 ) ^ { 2 } } = a \sqrt { \left( t ^ { 2 } - 1 \right) ^ { 2 } + 4 t ^ { 2 } }{/tex}= a(t2 + 1)
SQ = {tex}\sqrt { \left( \frac { a } { t ^ { 2 } } - a \right) ^ { 2 } + \left( \frac { 2 a } { t } - 0 \right) ^ { 2 } }{/tex}
SQ = {tex}\sqrt { \frac { a ^ { 2 } \left( 1 - t ^ { 2 } \right) ^ { 2 } } { t ^ { 4 } } + \frac { 4 a ^ { 2 } } { t ^ { 2 } } }{/tex}= {tex}\frac { a } { t ^ { 2 } } \sqrt { \left( 1 - t ^ { 2 } \right) ^ { 2 } + 4 t ^ { 2 } } = \frac { a } { t ^ { 2 } } \sqrt { \left( 1 + t ^ { 2 } \right) ^ { 2 } } = \frac { a } { t ^ { 2 } }( 1+ t^2){/tex}
{tex}\therefore \quad \frac { 1 } { S P } + \frac { 1 } { S Q } = \frac { 1 } { a \left( t ^ { 2 } + 1 \right) } + \frac { t ^ { 2 } } { a \left( t ^ { 2 } + 1 \right) }{/tex}
{tex}\Rightarrow \quad \frac { 1 } { S P } + \frac { 1 } { S Q } = \frac { 1 + t ^ { 2 } } { a \left( t ^ { 2 } + 1 \right) } = \frac { 1 } { a }{/tex}, which is independent of t.
Posted by Gokulan Gaffer 6 years, 10 months ago
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Ram Kushwah 6 years, 10 months ago
<i>whenever you are required to find a number that is </i><i>divisible</i><i> by more than one number, you have to find LCM.</i>
<i>whenever you are required to find a number that </i><i>completely divides</i><i> more than one number, you have to find HCF.</i>
Coming to illustrative examples :
<i>LCM</i>
<i>Three traffic signals change from Red to Green in 10, 15 & 20 seconds respectively. After how much time will all three signals together become Green ?</i>
The three traffic signals will change from Red to Green as follows ( in seconds ):
First : 10,20,30,40,50,60,70,80 …..
Second : 15,30,45,60,75,90 …
Third : 20,40,60,80 …
It is clear the three signals will flash Green together after 60 seconds. Now , this 60 is completely <i>divisible</i> by 10, 15 & 20 and LCM(10, 15 , 20) is 60, hence the problem was of finding LCM.
<i>HCF</i>
<i>Square towels has to be cut from a piece of cloth measuring 16m x 20m. What is the minimum number of towels that can be cut so that there is no wastage ?</i>
The towels are square and length is equal to width. Since there should be no wastage, the edge of the towel should exactly divide the length & breadth of the piece of cloth. The dimension of the towel is that highest number which completely divides 16 & 20. So, here we have a case of HCF.
HCF(16,20) = 4
So, the dimension of towel is 4m x 4m
The minimum number of towels possible is (16 x 20 )/ (4 x 4) = 20.
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Gaurav Seth 6 years, 10 months ago
Explanation: Use one of the Pythagorean identity namely,
sec2A=1+tan2A
sec4A−sec2A
=(sec2A)2−sec2A
=(1+tan2A)2−(1+tan2A)
=1+2tan2A+tan4A−1−tan2A
=tan4A+tan2A
2Thank You