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# 17thKVS-JMO2014-Solution3

## Probability Introduction :

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:

Given a = sin x + sin y   and b = cos x + cos y , using sin sum rule and cos sum rule we get ,

a = 2 $$\sin \left ( \frac{x+y}{2} \right )\cos \left ( \frac{x-y}{2} \right )$$

b = 2 $$\cos \left ( \frac{x+y}{2} \right )\cos \left ( \frac{x-y}{2} \right )$$

Given equation :

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

Divide the equation by   a² + b² + 2b  we get

t² - $$\large \frac{4a}{a^{2} + b^{2} + 2b }$$ t + $$\large \frac{a^{2} + b^{2} - 2b}{a^{2} + b^{2} + 2b }$$ = 0 .

Sum of the roots  = $$\large \frac{4a}{a^{2} + b^{2} + 2b}$$

Product of the roots = $$\large \frac{a^{2} + b^{2} -2b}{a^{2} + b^{2} +2b}$$

Now to prove that sum of the roots = $$\large tan (\frac{x}{2}$$) + tan($$\large \frac{y}{2}$$)

(OR)  Product of the roots = $$\large tan(\frac{x}{2}) * tan(\frac{y}{2})$$ .

Let us find the value of  :

a² + b² + 2b = 4 $$sin^{2} \left( \large \frac{x+y}{2} \right )$$ $$cos^{2}( \large \frac{x-y}{2} )$$ + 4 $$cos^{2} ( \large \frac{x+y}{2} )$$ $$cos^{2} ( \large \frac{x-y}{2}$$ )  +  4 cos$$( \large \frac{x+y}{2} )cos ( \large \frac{x-y}{2} )$$

= 4 $$\large \cos ^{2}\left ( \frac{x-y}{2} \right ) \left ( \sin ^{2} \left ( \frac{x+y}{2} \right) + \cos ^{2}\left ( \frac{x+y}{2} \right )\right )$$ + 4 $$\cos \left ( \frac{x+y}{2} \right )\cos \left ( \frac{x-y}{2} \right )$$

= 4 $$\large \cos \left ( \frac{x-y}{2} \right )\left ( \cos \left ( \frac{x-y}{2} \right )+\cos \left ( \frac{x+y}{2} \right ) \right )$$
= 4 $$\large\cos \left ( \frac{x-y}{2} \right )\left [ 2\cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right ) \right ]$$

= 8 $$\large\cos \left ( \frac{x-y}{2} \right ) \cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )$$

Sum of the roots =  $$\LARGE \frac{4a}{a^{2} + b^{2} + 2b}$$

= $$\large\frac {4\left ( 2\sin \left ( \frac{x+y}{2} \right )\cos \left ( \frac{x-y}{2} \right ) \right )}{8 \cos \left ( \frac{x-y}{2} \right )\cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )}$$

= $$\large \frac { \sin \left (\frac{x+y}{2} \right ) }{\cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )}$$

= $$\large \frac { \sin \left (\frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )+\cos \left ( \frac{x}{2} \right )\sin \left ( \frac{y}{2} \right ) }{\cos \left ( \frac{x}{2} \right )\cos \left ( \frac{y}{2} \right )}$$

= $$\large \tan \left ( \frac{x}{2} \right )+\tan \left ( \frac{y}{2} \right )$$

Therefore sum of the roots = $$\large \tan\left(\frac{x} {2} \right) + \tan \left (\frac {y} {2} \right$$

Similarly we prove product of roots = tan(x/2)*tan(y/2).

Hence tan(x/2) and tan(y/2) are roots of given equation .