Sunday 28 July 2019

#LDE of second order # method of variations of parameter

#LDE of Second Order # Solutions by means of operational factors

#LDE 2nd Order # Removal of first Derivative # Normal form

Jacobians

domain of functions

domain and range of functions by graphs

variables in functions

Properties of Addition of Vectors

Rectangular Resoultion of Vector and Direction Cosines

Basic Concept of Vectors

maxima and minima of two variables

Lagrange method of undetermined multiplier

Green Theorem

Thursday 16 May 2019

Important Short Tricks on Trigonometric indenties



Important Short Tricks on Trigonometric.  Indenties


Pythagorean Identities


sin2 Î¸ + cos2 Î¸ = 1


tan2 Î¸ + 1 = sec2 Î¸


cot2 Î¸ + 1 = csc2 Î¸


Negative of a Function


sin (–x) = –sin x


cos (–x) = cos x


tan (–x) = –tan x


csc (–x) = –csc x


sec (–x) = sec x


cot (–x) = –cot x


If A + B = 90o, Then


Sin A = Cos B


Sin2A + Sin2B = Cos2A + Cos2B = 1


Tan A = Cot B


Sec A = Csc B


For example:    


If tan (x+y) tan (x-y) = 1, then find tan (2x/3)?


Solution:            


Tan A = Cot B, Tan A*Tan B = 1


So, A +B = 90o


(x+y)+(x-y) = 90o, 2x = 90o , x = 45o


Tan (2x/3) = tan 30o = 1/√3


If A - B = 90o, (A › B) Then


Sin A = Cos B


Cos A = - Sin B


Tan A = - Cot B


If A ± B = 180o, then


Sin A = Sin B


Cos A = - Cos B


If A + B = 180o                   


Then, tan A = - tan B


If A - B = 180o                    


Then, tan A = tan B


 


If A + B + C = 180o, then


Tan A + Tan B +Tan C = Tan A * Tan B *Tan C


sin θ * sin 2θ * sin 4θ = ¼ sin 3θ


cos θ * cos 2θ * cos 4θ = ¼ cos 3θ


For Example:What is the value of cos 20ocos 40o cos 60o cos 80o?


Solution: We know cos θ * cos 2θ * cos 4θ = ¼ cos 3θ


Now, (cos 20o cos 40o cos 80o ) cos 60o


¼ (Cos 3*20) * cos 60o


¼ Cos2 60= ¼ * (½)2 = 1/16


If             a sin θ + b cos θ = m     &    a cos θ - b sin θ = n


then a2 + b2 = m2 + n2


For Example:


If 4 sin θ + 3 cos θ = 2 , then find the value of  4 cos θ - 3 sin θ:


Solution:


Let 2 cos θ - 3 sin θ = x


By using formulae a2 + b2 = m2 + n2


42 + 32 = 22 + x2


16 + 9 = 4 + x2


X = √21


If


sin θ +  cos θ = p     &     csc θ -  sec θ = q


then P – (1/p) = 2/q


For Example:


If sin θ + cos θ = 2 , then find the value of  csc θ - sec θ:


Solution:


By using formulae:


P – (1/p) = 2/q


2-(1/2) = 3/2 = 2/q


Q = 4/3 or csc θ - sec θ = 4/3


If


a cot θ + b csc θ = m     &    a csc θ + b cot θ = n


then b2 - a2  = m2 - n2


If


cot θ + cos θ = x     &    cot θ - cos θ = y


then x2 - y2 = 4 √xy


If


tan θ + sin θ = x     &    tan θ - sin θ = y


then x2 - y2 = 4 √xy


If


y = a2 sin2x + b2 csc2x + c


y = a2 cos2x + b2 sec2x + c


y = a2 tan2x + b2 cot2x + c


then,


ymin = 2ab + c


ymax = not defined


For Example:                    


If y = 9 sin2 x + 16 csc2 x +4 then ymin is:


Solution:            


For, y min = 2* √9 * √16 + 4


= 2*3*4 + 20 = 24 + 4 = 28


If            


y = a sin x + b cos x + c


y = a tan x + b cot x + c


y = a sec x + b csc x + c


then,     ymin = + [√(a2+b2)] + c


ymax = - [√(a2+b2)] + c


For Example:                    


If y = 1/(12sin x + 5 cos x +20) then ymaxis:


Solution:            


For, y max = 1/x min


= 1/- (√122 +52) +20 = 1/(-13+20) = 1/7


Sin2 Î¸, maxima value = 1, minima value = 0


Cos2 Î¸, maxima value = 1, minima value = 0




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