IIT JEE , AIEEE Forum - Mathematics , Trignometry - given:a+b+c=180;tan{(a+b-c)/4}tan{(b+c-a)/4}tan{(a+c-b)/4}=1RTP-1+cosa+cosb+cosc=

yoganda

given:a+b+c=180; & tan{(a+b-c)/4}tan{(b+c-a) /4}tan{(a+c-b)/4}=1

RTP-1+ cosa+cosb+ cosc= 0

ramkumar_november

Let x=tan a/2 , y=tan b/2 , z=tan c/2

a+b+c =

a/2 + b/2 + c/2 =

/2

tan(a/2 + b/2 + c/2)= infinity

(S₁ - S₃)/(1-S₂)=INFINITY

=> S₂=1

=> xy + yz + xz = 1.......................(i)

Given
tan{(a+b-c)/4}tan{(b+c-a)/4}tan{(a+c-b)/4}=1

tan{

/4 - c/2} tan{

/4 - b/2} tan{

/4 - a/2} = 1

=> (1-x)(1-y)(1-z)
      ---------------------        =      1
     (1+x)(1+y)(1+z)

On cross multiplying and expanding we get,

=> x + y + z = -xyz..................(ii)

LHS

= 1 + cosa + cosb + cosc

= 1+ (1-x²)/(1+x²) + (1-y²)/(1+y²) + (1-z²)/(1+z²)

= 1 - x²   + 1 + 1 - y² + 1 + 1 - z² + 1 - 2
   ---------              --------     ---------
    1+x²                1+y²              1+z²

=     2      +       2     +      2       -      2
   ----------      ----------    -----------
     1+x²        1 + y²      1 +z²

= 2 { (1+y²)(1+z²) + (1+x²)(1+y²) + (1+x²)(1+z²) - (1+x²)(1+y²)(1+z²) }
         --------------------------------------------------------------------------------------------------
                                    (1+x²)(1+y²)(1+z²)

= 2 { 2+ (x²+y²+z² ) + x²y² +y²z² + x²z²   - ( x²y² +y²z² + x²z² ) - x²y²z² }
        -------------------------------------------------------------------------------------------------------
                             (1+x²)(1+y²)(1+z²)

= 2 { 2 + (x+y+z)² - 2(xy+yz+xz) - x²y²z²}
        -----------------------------------------------------
            (1+x²)(1+y²)(1+z²)

= 2 { 2 + x²y²z² - 2(1) - x²y²z² }
   -----------------------------------------
       (1+x²)(1+y²)(1+z²)

= 2 (0)

=0

=RHS

hence 1 + cosa +cosb +cosc =0...................

please rate me if you are satisfied with my proof...........................

bvsatyaram

Good Ramkumar!!

The technique used in this solution is called, method of universal substitution,
x=tan (x/2) is the universal subtitution we are tallking abt.
using this substitution, many complicated trignometric problems can be converted to algebraic problems.
Once we express sinx and cosx in terms of tan (x/2), the whole expression turns to a an expression in tan (x/2). Which will be purely algebraic, once we put t=tan (x/2).

ayshwarya

hey but in exam u wud suffer by doing dees medod ive done in anoder pattern see dis
given tan[pie/4-2c/4]tan[pie/4-2b/4]tanpie/4-2a/4]=1

hence by putting a=0 b=0 c=pie [even pie/2 , pie/2, 0 b chalta]

hence v wud get -1+1-1+1=0 hence proved rate me if u feel good

harsha_27

substituting a=0 , b=0 , c = pie in

tan{

/4 - c/2} tan{

/4 - b/2} tan{

/4 - a/2} = 1

gives tan{-

/4}tan{

/4} = -1

1

nandinis_7

hey under what conditions can v use this method of universal substitution?

bvsatyaram

There is no specific condition for using universal substitution. If the problem becomes very tough to solve using pure trigonometric concepts, and if u feel that converting the trigonometric equation into an arithmetic one helps, then u use the methods of universal substitution.

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