1.If x=∑n=0∞(−1)ntan2nθx = \sum\limits_{n=0}^{\infty} (-1)^n \tan^{2n} \thetax=n=0∑∞(−1)ntan2nθ and y=∑n=0∞cos2nθy = \sum\limits_{n=0}^{\infty} \cos^{2n} \thetay=n=0∑∞cos2nθ, for 0<θ<π40 < \theta < \dfrac{\pi}{4}0<θ<4π, thena.x(1+y)=1x(1 + y) = 1x(1+y)=1b.y(1−x)=1y(1 - x) = 1y(1−x)=1c.y(1+x)=1y(1 + x) = 1y(1+x)=1d.x(1−y)=1x(1 - y) = 1x(1−y)=1Login to continueOnly logged in users canattempt or see the solution.
