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17thKVS-JMO2014-Solutions1

17thKVS-JMO2014-Solutions1

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 3Q) Let sin x + sin y = a and cos x + cos y = b, Show that tan\(\large \frac{x}{2}\) + tan\(\large \frac{y}{2}\) are two roots of the equation:

                              (a² + b² + 2b)t² - 4at + (a² + b² - 2b) = 0

Solution :

                              (a² + b² + 2b)t² - 4at + (a² + b² - 2b) = 0

   Product of the roots = \(\large \frac{a² + b² - 2b}{a² + b² + 2b}\)    -----------(1)
   Sum of the roots     = \(\large\frac{4a}{a² + b² + 2b}\)

 Now let us assume that tan\(\large \frac{x}{2}\) and tan\(\large \frac{y}{2}\) are the roots and then prove that
              tan\(\large \frac{x}{2}\)  +  tan\(\large \frac{y}{2}\)  =  \(\large \frac{4a}{a² + b² + 2b}\).

 We know that,
      tan\(\large \frac{x}{2}\)  +  tan\(\large \frac{y}{2}\)  =  tan ( \(\large \frac{x}{2} + \frac{y}{2}\) ) ( 1 - tan\(\large \frac{x}{2}\) tan\(\large \frac{y}{2}\) )                   ---------------(2)

 We have,
      a = sin x  +  sin y
      b = cos x  +  cos y
 

                       \(\large \frac{a}{b}\) = \(\large \frac{sin x + sin y}{cos x + cos y}\)

                            = \(\large \frac{2sin(\frac{x+y}{2})cos(\frac{x-y}{2})}{2cos(\frac{x+y}{2})cos(\frac{x-y}{2})}\)

                            = tan(\(\large \frac{x+y}{2}\))          ------------(3)

Plug equations (1) & (3) in equation (2), we get,

        tan( \(\large \frac{x}{2}\) ) + tan (\(\large \frac{y}{2}\) ) = \(\large \frac{a}{b}\) \(\large (1 - \frac{a² + b² - 2b}{a² + b² + 2b}) \) 

                                        = \(\large \frac{a}{b} ( \frac{a² + b² + 2b - a² - b² + 2b}{a² + b² + 2b} ) \)

                                        = \(\large \frac{a}{b} ( \frac{4b}{a² + b² + 2b}) \)     

                                        = \(\large \frac{4a}{a² + b² + 2b} \)

                                        = \(\large \alpha + \beta\)

  Therefore tan( \(\large \frac{x}{2} \) ) and tan ( \(\large \frac{y}{2}\) ) are the roots of the given equation.

         We can solve this in another method. For this click (OR) link below.

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