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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.