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# Let α and β be non–zero real numbers such that 2(cosβ – cosα) + cosα cosβ = 1. Then which of the following is/are true?

Umang Kumar
15 Points
2 years ago
2(cosb-cosa)+cosa.cosb=1
2cosb-2cosa+cosa.cosb=1
cosb(2+cosa)-(2cosa+1)=0
$cos\beta =\frac{2cos\alpha +1}{2+cos\alpha }$
Now converting $cos\beta =\frac{1-tan^{2}\frac{\beta }{2}}{1+tan^{2}\frac{\beta }{2}}$
$cos\alpha =\frac{1-tan^{2}\frac{\alpha }{2}}{1+tan^{2}\frac{\alpha }{2}}$
now equating cosa and cosb in
$cos\beta =\frac{2cos\alpha +1}{2+cos\alpha }$
we get
$\frac{1-tan^{2}\frac{\beta }{2}}{1+tan^{2}\frac{\beta }{2}}=\frac{3-tan^{2}\frac{\alpha }{2}}{3+tan^{2}\frac{\alpha }{2}}$
After applying Componendo and dividendo method each side we get:
$\frac{2}{-2tan^{2}\frac{\beta }{2}}=\frac{6}{-2tan^{2}\frac{\alpha }{2}}$
After solving this we get:
$\frac{1}{tan^{2}\frac{\beta }{2}}=\frac{3}{tan^{2}\frac{\alpha }{2}}$
After cross multiplying we get:
$tan^{2\frac{\alpha }{2}}=3tan^{2\frac{\beta }{2}}$
Square root both side:
$tan\frac{\alpha }{2}=\pm \sqrt{3}tan\frac{\beta }{2}$
hence the answers are:-
1.$tan\frac{\alpha }{2}+\sqrt{3}tan\frac{\beta }{2}=0$
2.$tan\frac{\alpha }{2}-\sqrt{3}tan\frac{\beta }{2}=0$