JEE MAIN - Mathematics (2019 - 11th January Evening Slot - No. 11)
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 :
$${1 \over {1 + \cos \theta }} + {1 \over {1 - \sin \theta }}$$
$${1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$$
$${1 \over {1 - \cos \theta }} - {1 \over {1 + \sin \theta }}$$
$${1 \over {1 + \cos \theta }} - {1 \over {1 - \sin \theta }}$$
Explanation
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$$\theta $$
$$\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{\left( { - {1 \over \beta }} \right)}^n}} \right)} = \sum\limits_{n = 0}^\infty {{{\left( {\cos \theta } \right)}^n}} + \sum\limits_{n = 0}^n {{{\left( { - \sin \theta } \right)}^n}} $$
$$ = {1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$$
$$ \Rightarrow $$ roots are $$\beta $$ = cosec$$\theta $$ and $$\alpha $$ = cos$$\theta $$
$$\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{\left( { - {1 \over \beta }} \right)}^n}} \right)} = \sum\limits_{n = 0}^\infty {{{\left( {\cos \theta } \right)}^n}} + \sum\limits_{n = 0}^n {{{\left( { - \sin \theta } \right)}^n}} $$
$$ = {1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$$
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