JEE MAIN - Mathematics (2020 - 9th January Evening Slot - No. 11)
If $$x = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta } $$ and $$y = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } $$
for 0 < $$\theta $$ < $${\pi \over 4}$$, then :
for 0 < $$\theta $$ < $${\pi \over 4}$$, then :
x(1 + y) = 1
y(1 – x) = 1
y(1 + x) = 1
x(1 – y) = 1
Explanation
$$x = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta } $$
= 1 – tan2$$\theta $$ + tan2 4$$\theta $$ + ...
= $${1 \over {1 + {{\tan }^2}\theta }}$$ = cos2 $$\theta $$ ....(1)
$$y = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } $$
= 1 + cos2 $$\theta $$ + cos4 $$\theta $$ + cos6 $$\theta $$ + ....
= $${1 \over {1 - {{\cos }^2}\theta }}$$ = $${1 \over {{{\sin }^2}\theta }}$$
$$ \Rightarrow $$ sin2 $$\theta $$ = $${1 \over y}$$ ...(2)
Adding (1) and (2), we get,
x + $${1 \over y}$$ = sin2 $$\theta $$ + cos2 $$\theta $$
$$ \Rightarrow $$ x + $${1 \over y}$$ = 1
$$ \Rightarrow $$ y(1 – x) = 1
= 1 – tan2$$\theta $$ + tan2 4$$\theta $$ + ...
= $${1 \over {1 + {{\tan }^2}\theta }}$$ = cos2 $$\theta $$ ....(1)
$$y = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta } $$
= 1 + cos2 $$\theta $$ + cos4 $$\theta $$ + cos6 $$\theta $$ + ....
= $${1 \over {1 - {{\cos }^2}\theta }}$$ = $${1 \over {{{\sin }^2}\theta }}$$
$$ \Rightarrow $$ sin2 $$\theta $$ = $${1 \over y}$$ ...(2)
Adding (1) and (2), we get,
x + $${1 \over y}$$ = sin2 $$\theta $$ + cos2 $$\theta $$
$$ \Rightarrow $$ x + $${1 \over y}$$ = 1
$$ \Rightarrow $$ y(1 – x) = 1
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