Let [ ] and [ ] be the roots of the quadratic equation x2 sin [ ] -- x(sin [ ] cos [ ] + 1…

Question

Let $$\alpha $$ and $$\beta $$ be the roots of the quadratic equation x2
 sin $$\theta $$ -- x(sin $$\theta $$ cos $$\theta $$ + 1) + cos $$\theta $$ = 0 (0 < $$\theta $$ < 45o), and $$\alpha $$ < $$\beta $$. Then $$\sum\limits_{n = 0}^\infty  {\left( {{\alpha ^n} + {{{{\left( { - 1} \right)}^n}} \over {{\beta ^n}}}} \right)} $$ is equal to :

Accepted Answer

D = (1 + sin$$\theta $$ cos$$\theta $$)2 - 4sin$$\theta $$cos$$\theta $$ = (1 - sin$$\theta $$ cos$$\theta $$)2

$$ \Rightarrow $$  roots are $$\beta $$ = cosec$$\theta $$ and $$\alpha $$ = cos$$\theta $$

$$\sum\limits_{n = 0}^\infty  {\left( {{\alpha ^n} + {{\left( { - {1 \over \beta }} \right)}^n}} \right)}  = \sum\limits_{n = 0}^\infty  {{{\left( {\cos \theta } \right)}^n}}  + \sum\limits_{n = 0}^n {{{\left( { - \sin \theta } \right)}^n}} $$

$$ = {1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$$

Solutions · 1