let z = x + iy

          z = (m-in)/(1-l) =  x + iy 

or      x + iy  = ( m / (1-l)) + ( - in / (1-l) )

        compare both side--

or     x = m / ( 1-l )     and   y  = n / ( l -1)

L.H.S-   ( 1-iz /1+iz ) = ( 1 - (x + iy  ) ) / ( 1 + (x + iy  ) ) 

    or

      -----------------------Eq.(1)

put the value x and y -- partially solve numerator and denomenator

part(1)-

   put the value (  m²+n²=1 - l²  )

part-(2)

put the value of part(1) and part(2) in Eq.(1)---

prove this.......

well try  vnkt.swaroop ,

Dear REJAULMA,

I think a,b,c are the sides of triangle and Let A,B,C are the respective angles which are in AP

so 2B =A+C            and since A+B+C =180

hence 2B = 180-B

hence B=60

now apply Sin Rule:

SinB/b =Sin C/c

or Sin60 /b = Sin C/c

or sinC = c Sin60 /b

or sin² C = c² Sin²60 /b²  =3c²  /4b2

from given eq:  sin² C = 2b²/4b²

or sin² C = 1/2   or Sin C =1 / root(2)    =>   C =45

hence ,A = 75

good work sboosy

well done

Dear,

Let P be (x,y).

in triangle APC , LET <APC = A  and < PCA = B

 hence ,<BPC = A    and <PCB = (180-B)

Apply sin  Rule  in both triangles:

(Sin A / AC ) = (Sin B/ AP)  ..................eq(1)

and (Sin A / CB ) = (Sin B/ PB).................eq(2)

from (1) and (2) , equate Sin A/Sin B:

(AC / AP) = (CB / PB)

or (AC / CB) = (AP / PB)

or tan = (AP / PB)                             since (AC/CB) = tan

or PB tan = AP

or PB² tan² = AP²

or [(x-a)²+y² ] tan² = [(x+a)²+y² ]

or x²(tan² -1)+y² (tan² -1) + a² (tan² -1) - 2ax (tan² +1)=0

or x²+y²+a² - 2ax (tan² +1)/(tan² -1)=0  

or x²+y²+a² + 2ax (tan² +1)/(1-tan²)=0  

or x²+y²+a² + 2ax (Sin² +Cos²)/(Cos²-Sin²)=0   

or x²+y²+a² + 2ax (1)/(Cos2)=0  

or x²+y²+a² + 2ax Sec2=0

please check your question it may be sec instead of cosec.

Dear

 the condition is valid only when the slopes have some defined  values.

but in this case since slope of y axis is undefined so the condition is not valid for this case and all other cases where the lines are parellel  to  y axis e.g. x=2 or so on........

so flaringstorm is right.

good work  sachinguptaiit

let I =sec(3x)sec(5x)sec(8x)

       = 1/(cos(3x)cos(5x)cos(8x)) =  ( A/cos(3x)) + ( B/cos(5x)) + ( C/cos(8x))  ----------Eq.(Q)

   ==>>>    1 = Acos(5x)cos(8x) + Bcos(3x)cos(8x) +  Ccos(3x)cos(5x)         ----------------- Eq.(1)

             in Eq.(1)

         component of A and B will be zero bcoz cos(pie by 2)= 0;

  then    

                  in Eq.(1)

              component of A and C will be zero bcoz cos(pie by 2)= 0;

          in Eq.(1)

         component of B and C will be zero bcoz cos(pie by 2)= 0;

        PUT THE VALUE  A,B  AND  C in eq.(Q)  

       Integrate..... 

put the value

let base a=6, differ. of base angles B-C=60⁰,  AREA=12

                   A+B+C=180⁰

                    B+C=180⁰ - A

               TAKEN Cos both side---

                  Cos(B+C)=Cos(180⁰ + A )

                  Cos(B)Cos(C) - Sin(B)Sin(C) = - Cos(A)  ---------------------  Eq.(1)

                  B - C =60⁰

                TAKEN Cos both side---

               Cos(B - C)=Cos(60⁰)

               Cos(B)Cos(C) + Sin(B)Sin(C) = 1/2 ----------------------------Eq.(2)

                Eq.(2) - Eq(1)

               2Sin(B)Sin(C) = Cos(A) + (1/2)  ----------------------Eq.(3)

               As we know that--

            put the value SinB and SinC in Eq.(3)

          Hence Prove.....

=

Applying R₁-->aR₁ , R₂-->bR₂ and R₃-->cR₃

Applying C_3--> C_{1 +} C₃

taking (ab+ac+bc) common

Applying C_3--> C_{3 -} C₂

let A = Tanx+Cotx+Secx+Cosecx

     A = (Sinx/Cosx)+(Cosx/Sinx)+(1/Cosx)+(1/Sinx)

        = (Sinx^2+Cosx^2+Sinx+Cosx)/Sinx.Cosx

        = (1 + Sinx+Cosx)/Sinx.Cosx

 multiply by (Sinx+Cosx-1) in denominator and numerator..     

     A  = (Sinx+Cosx+1)(Sinx+Cosx-1) / (Sinx.Cosx)(Sinx+Cosx-1)

          = (Sinx^2+Cosx^2+2Sinx.Cosx-1) / (Sinx.Cosx)(Sinx+Cosx-1)

          = (1+2Sinx.Cosx-1) / (Sinx.Cosx)(Sinx+Cosx-1)     

          = (2Sinx.Cosx) / (Sinx.Cosx)(Sinx+Cosx-1)    

          = 2 / (Sinx+Cosx-1)  

    I     = integrate dx/(Tanx+Cotx+Secx+Cosecx)

    I    = integrate dx/A

    I    = integrate (Sinx+Cosx-1) dx/2

    I   =  (-Cosx + Sinx -x)/2