Verified Solution • 2025–26 Session moderate

QJEEM20Ma547e: JEE Main 2020 Mathematics - Class 11 JEEM20Ma547e Solution: If and for 0 < < , then :

If x=∑n=0∞(−1)ntan⁡2nθx = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta } and y=∑n=0∞cos⁡2nθy = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta }

for 0 < θ\theta < π4{\pi \over 4} , then :

Multiple Choice Options

  • A x(1 + y) = 1
  • B y(1 – x) = 1
  • C y(1 + x) = 1
  • D x(1 – y) = 1

Step-by-Step Verified Solution

x=∑n=0∞(−1)ntan⁡2nθx = \sum\limits_{n = 0}^\infty {{{\left( { - 1} \right)}^n}{{\tan }^{2n}}\theta }

= 1 – tan2 θ\theta + tan2 4 θ\theta + ...

= 11+tan⁡2θ{1 \over {1 + {{\tan }^2}\theta }} = cos2 θ\theta ....(1)

y=∑n=0∞cos⁡2nθy = \sum\limits_{n = 0}^\infty {{{\cos }^{2n}}\theta }

= 1 + cos2 θ\theta + cos4 θ\theta + cos6 θ\theta + ....

= 11−cos⁡2θ{1 \over {1 - {{\cos }^2}\theta }} = 1sin⁡2θ{1 \over {{{\sin }^2}\theta }}

⇒\Rightarrow sin2 θ\theta = 1y{1 \over y} ...(2)

Adding (1) and (2), we get,

x + 1y{1 \over y} = sin2 θ\theta + cos2 θ\theta

⇒\Rightarrow x + 1y{1 \over y} = 1

⇒\Rightarrow y(1 – x) = 1

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