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True Discount and Banker's Discount problems tricks in Hindi | fast track formulae for problem solving.


TRUE DISCOUNT 1). वास्तविक बट्टा = मिश्रधन - वर्तमान धन
2). यदि ब्याज की दर r% वार्षिक , समय t व वर्तमान धन (pw) है, तो वास्तविक बट्टा = PW×r×t/100
3). यदि r% तथा समय t के बाद देय धन A है, तो तत्काल धन pw = 100×A/(100+r.t)
4). यदि देय धन A पर r% व समय t दिए गए है, तो वास्तविक बट्टा = A.r.t /(100+r.t)
5). यदि किसी निश्चित समय के पश्चात निश्चित वार्षिक दर पर, देय धन पर वास्तविक बट्टा (TD) व समान समय व दर के लिए साधारण ब्याज (SI) हैं, तो देय धन A =      SI × TD/(SI - TD)
6). यदि t समय पश्चात r% वार्षिक दर पर, देय धन पर वास्तविक बट्टा (TD) व साधारण ब्याज (SI) है , तो -  SI - TD = TD×r×t/100
7). t वर्ष बाद r% चक्रवृद्धि दर पर देय धन A का तत्काल धन (PW) = A/(1+r/100)^t   व वास्तविक बट्टा = A - PW
BANKER'S DISCOUNT1). महाजनी बट्टा = शेष समय (समाप्त न हुए समय) के लिए बिल पर ब्याज = बिल की राशि × दर × शेष समय /100
2). महाजनी लाभ = महाजनी बट्टा - वास्तविक बट्टा
3). यदि बिल का मान / अंकित मूल्य A है, समय t व दर r% है, तो महाजनी बट्टा = A×r×t/100

Inverse trigonometrc function IIT JEE study material | ITF concept booster | ITF notes | ITF important formulae.



If f:X⟶ Y is a function which is both one - one and onto, then it's inverse function 
f^-1:Y⟶X is defined as: y = f(x) ⟺ f^-1(y)=x
Such that , ∀x∈X , ∀y∈Y


Let's take a sine function, whose domain is R and range [-1,1]. We see that this function is many -one and onto, so , it's inverse doesn't exist, but if we restrict the domain of the sine function to the interval [-π/2 , π/2] , then the function is:

sin: [-π/2 , π/2] ⟶  [-1,1] and given by sinθ = x
Which is one-one and onto and therefore it's inverse exist.
And the Inverse of sine function is defined as
Sin-1: [-1,1] ⟶ [-π/2 , π/2], such that sin-1x = θ
From this we can say that if x is a real numbers between -1 and 1 then θ will be in -π/2 and π/2.

Sin-1x = θ then x = sinθ , where, -π/2⪯ θ⪯  π/2 and -1 ⪯x⪯ 1
So , the least numerical value among all the values of the angle whose sine is x , is called the principal value of sin-1x. So we can give similar definition for cos-1x , tan-1x etc.

*. Graphs of inverse trigonometry function can be drawn from the knowledge of the graphs of the corresponding trigonometry function.
*. In ITF graphs can be obtained by interchanging X and Y axis.
The graphs of inverse trigonometry function are given below:
*. Keep in mind that if the domain of ITF is not stated then always consider principal value of given Inverse function.

sin-1x and (sinx)^-1 is different terms , they are not equal
∴ sin-1x ≠ sinx)^-1 , similarly for other functions.


1). *. sin-1(sinθ) = θ and sin(sin-1x) = x , provided -1 ⪯x⪯ 1 and -π/2⪯ θ⪯  π/2

*. cos-1(cosθ) = θ and cos(cos-1x) = x , provided -1 ⪯x⪯ 1 and  0⪯ θ⪯  π

*. tan-1(tanθ) = θ and tan(tan-1x) = x , provided -∞<x<∞ and -π/2<θ<π/2

*. cot-1(cotθ) = θ and cot(cot-1x) = x , provided -∞<x<∞ and 0<θ<π

*. sec-1(secθ) = θ and sec(sec-1x) = x
*. cosec-1(cosecθ) = θ and cosec(cosec-1x) = x

2). *. sin-1x = csc-1(1/x) or csc-1x = sin-1(1/x)
*. cos-1x = sec-1(1/x) or sec-1x = cos-1(1/x)

*. tan-1x = cot-1(1/x) if x>0
 and tan-1x = cot-1(1/x) -π if x<0
and cot-1x = tan-1(1/x) if x >0
 and cot-1x = tan-1(1/x) +π if x<0

3). *. Sin-1x = cos-1√(1- x²) = tan-1(x/√1-x²) = 
 = cot-1(√1-x²/x) = sec-1(1/√1-x²) = csc-1(1/x)

*. Cos-1x = sin-1(√1-x²) = tan-1(√1-x²/x) = 
  = Cot-1(x/√1-x²) = sec-1(1/x) = csc-1(1/√1-x²)

*. Tan-1x = sin-1(x/√1+x²) = cos-1(1/√1+x²)
   = Cot-1(1/x) = sec-1(√1+x²) = csc-1(√1+x²/x)

4). *. Sin-1x + cos-1x = π/2 , where -1 ⪯x⪯ 1
*. Tan-1x + cot-1x = π/2 , where -∞<x<∞
*. Sec-1x + csc-1x = π/2 , where x⪯ -1 or 1⪯x

5). *. Sin-1x +sin-1y = sin-1(x√1-y² + y√1-x²)
 If xy⪯0 or ( xy>0 and x² +y² ⪯ 1)

*. Sin-1x - sin-1y = sin-1(x√1-y² - y√1-x²)
 If 0⪯xy or ( xy<0 and x² +y² ⪯ 1)

*. Cos-1x + cos-1y = cos-1(xy - √1-y².√1-x²),. If 
  |x| , |y| ⪯ 1 , 0 ⪯ x+y

*. cos-1x - cos-1y = cos-1(xy + √1-y².√1-x²)
If |x| , |y| ⪯ 1, x ⪯ y

*. Tan-1x + tan-1y = tan-1(x+y/1-xy), if xy < 1
*. Tan-1x - tan-1y = tan-1(x-y/1+xy), if xy > -1

*. 2sin-1x = sin-1(2x√1- x²), if -1/√2 ⪯ x ⪯ 1/√2
*. 2cos-1x = cos-1(2x² - 1) ,if 0 ⪯ x ⪯ 1

*. 2tan-1x = tan-1(2x/1-x²) if -1<x<1
   2tan-1x = sin-1(2x/1+x²) , if -1 ⪯x⪯ 1
   = cos-1(1-x²/1+x²), if 0⪯x<∞

6). *. Sin-1(-x) = - sin-1x 
     *. Cos-1(-x) = π - cos-1x
     *. Tan-1(-x) = - tan-1x
     *. cot-1(-x) = π - cot-1x

7). *. 3sin-1x = sin-1( 3x - 4x^3), if -1/2 ⪯ x⪯ 1/2
      *. 3cos-1x = cos-1(4x^3 - 3x) , if 1/2 ⪯ x ⪯ 1
      *. 3tan-1x = tan-1[(3x - x^3)/(1 - 3x²)] , if 
          -1/√3 < x < 1/√3

8). *. tan-1x + tan-1y + tan-1z
          = tan-1[(x+y+z-xyz)/(1- xy - yz - zx)]

*. sin-1x , cos-1x , tan-1x can also be written as arc sinx , arc cosx, arc tanx.
*. If it's not stated then always consider principal value of the Inverse trigonometry function.

*. If tan-1x + tan-1y + tan-1z = π/2 , then xy + yz + zx = 1

*. tan-1x + tan-1y + tan-1z = π , then x+y+z = xyz

*. If sin-1x + sin-1y + sin-1z = π/2 , then 
   x^2 + y^2 + y^2 + 2xyz = 1 

 *. If sin-1x + sin-1y + sin-1z = π , then 
  x√1- x^2 + y√ 1-y^2 + z√1-z^2 = 2xyz

*. If cos-1x + cos-1y + cos-1z = 3π 
 then xy +yz +zx = 3

*. If cos-1x + cos-1y + cos-1z = π , then 
       x^2 + y^2 + y^2 + 2xyz = 1

*. If sin-1x + sin-1y + sin-1z = 3π/2 , then 
      xy +yz +zx = 3

*. If sin-1x + sin-1y = θ ,then cos-1x + cos-1y = π-θ
*. If cos-1x + cos-1y = θ ,then sin-1x + sin-1y= π-θ

*. If cos-1(x/a) + cos-1(y/b) = θ , then 
  (x/a)^2 + (y/b)^2 - 2xycosθ/ab = sin^2θ.



Boat and stream short-tricks in Hindi | Fast track arithmetic formulae for competitive examination.

BOAT AND STREAM (नाव एवं धारा)
1). यदि शांत जल में नाव या तैराक की चाल x किमी/घंटा व धारा की चाल y किमी/घंटा है, तो धारा के अनुकूल नाव अथवा तैराक की चाल = (x+y) किमी/घंटा
2). धारा के प्रतिकूल नाव अथवा तैराक की चाल = (x-y) किमी /घंटा
3). नाव की चाल = (अनुप्रवाह चाल + उद्धर्वप्रवाह चाल)/2
4). धारा की चाल =  (अनुप्रवाह चाल - उद्धर्वप्रवाह चाल)/2
5). यदि धारा की चाल a किमी/घंटा है, तथा किसी नाव अथवा तैराक को उद्धर्वप्रवाह जाने में अनुप्रवाह जाने के समय का n गुना समय लगता है,(समान दूरी के लिए), तो शांत जल में नाव की चाल = a(n+1)/(n-1) किमी/घंटा
6). शांत जल में किसी नाव की चाल x किमी/घंटा व धारा की चाल y किमी/घंटा है, यदि नाव द्वारा एक स्थान से दूसरे स्थान तक आने व जाने में T समय लगता है, तो दोनो स्थानों के बीच की दूरी = T(x^2 - y^2)/2x km
7). कोई नाव अनुप्रवाह में कोई दूरी a घंटे में तय करती है, तथा वापस आने में b घंटे लेती है, यदि नाव कि चाल c किमी/घंटा है, तो शांत जल में नाव की चाल = c(a+b)/(b-a) km/h
8). यदि शांत जल में नाव की चाल a किमी/घंटा है, तथा वह b किमी/घंटा की चाल से बहती हुई नदी में गत…

Emergence of British East India Company as an Imperialist Political Power in India

Dynamically changing India during early eighteenth century had a substantially growing economy under the authority of Mughal emperor Aurangzeb. But after his demise in 1707, several Mughal governors established their control over many regional kingdoms by exerting their authority. By the second half of eighteen century, British East India Company emerged as a political power in India after deposing regional powers and dominating over Mughal rulers. The present article attempts to analyze the reasons for emergence of British East India Company as an imperial political power in India and their diplomatic policies of territorial expansion. In addition to this, I briefly highlighted the Charter Acts (1773, 1793, 1813, 1833 and 1853) to trace its impact on the working process of Company.Establishment of East India Company in India

In 1600, British East India Company received royal charter or exclusive license…

CORONA VIRUS, history of origin , discovery , infection mechanism, symptoms and treatment.

Today we are going to talk about a virus , which is spreading very fastly all over the world. The virus which we are going to talk about is the CORONA VIRUS. So today we will talk about everything of this virus. So let's starts ...

According to the biological study , Coronavirus is a cluster of viruses that causes diseases in birds and mammals. Therefore humans are also mammals then in human being this viruses cause respiratory infections , and one of the respiratory infections is mild common cold. Coronavirus can lead to diarrhea in cows and pigs but in chicken they can cause upper respiratory infections. Currently there is no vaccine or antiviral drugs for the treatment of diseases caused by Coronavirus.
The family of Coronavirus is coronaviridae, and it's subfamily is Orthocoronavirinae and order is Nidovirales, Coronavirus is a member of Orthocoronavirinae subfamily. All Coronavirus is coated with positive sense single …

Three dimensional geometry (part-1) | study material for IIT JEE | concept booster , chapter highlights.


In the following diagram X'OX , Y'OY and Z'OZ are three mutually perpendicular lines , which intersect at point O. Then the point O is called origin.
In the above diagram X'OX is called the X axes, Y'OY is called the Y axes and Z'OZ is called the Z axes.
1). XOY is called the XY plane. 2). YOZ is called the YZ plane. 3). ZOX is called the ZX plane.
If all these three are taken together then it is called the coordinate planes. These coordinates planes divides the space into 8 parts and these parts are called octants.
COORDINATES  Let's take a any point P in the space. Draw PL , PM and PN perpendicularly to the XY, YZ and ZX planes, then
1). LP is called the X - coordinate of point P. 2). MP is called the Y - coordinate of point P. 3). NP is called the Z - coordinate of the point P.
When these three coordinates are taken together, then it is called coordinates of the point P.

Speed , Distance and Time problems tricks in Hindi | fast track arithmetic formulae for problem solving.

1). दूरी = चाल × समय
2). समय = दूरी/चाल
3). चाल = दूरी/समय
4). किलोमीटर को मील बनाने के लिए गुना किया जाता है =       5/8 से
5). मील को किलोमीटर बनाने के लिए गुना किया जाता है =       8/5 से
6). फुट - सेकंड को मील - घंटा बनाने के लिए गुना किया जाता है = 15/22 से
7). मील - घंटा को फुट - सेकंड बनाने के लिए गुना किया जाता है = 22/15 से
8). मी - सेकंड को किमी - घंटा बनाने के लिए गुना किया जाता है = 18/5 से
9). किमी - घंटा को मी - सेकंड बनाने के लिए गुना किया जाता है = 5/18 से
10). यदि एक व्यक्ति दो निश्चित स्थानों के बीच की दूरी a किमी/घंटा की चाल से खत्म करता है, तो t1 घंटे देर से पहुंचता है, तथा जब b किमी/घंटा की चाल से तय करता है, तब वह t2 घण्टे पहले पहुंचता है, तो दोनो स्थानों के बीच की दूरी =     ab(t1+t2)/(b-a) km
11). यदि कोई व्यक्ति a km/h की चाल से चलता है, तो वह अपनी मंजिल पर t1 घंटे लेट पहुंचता है, अगली बार वह अपनी चाल में b km/h की वृद्धि करता है, तो वह t2 घंटे लेट पहुंचता है, तब उसके द्वारा तय की गई दूरी = a(a+b)(t1-t2)/b
12). दो व्यक्ति X …