Solveeit Logo
Login

Question

Question: Given that 0 \< x \<\(\frac{\pi}{4}\),\(\frac{\pi}{4}\)\< y \<\(\frac{\pi}{2}\) and \(\sum_{k = 0}^{...

Given that 0 < x <π4\frac{\pi}{4},π4\frac{\pi}{4}< y <π2\frac{\pi}{2} and ∑k=0∞(–1)k\sum_{k = 0}^{\infty}{(–1)^{k}}tan2k x = p, ∑k=0∞(–1)k\sum_{k = 0}^{\infty}{(–1)^{k}}cot2k y = q, then ∑k=0∞tan⁡2k\sum_{k = 0}^{\infty}\tan^{2k}x cot2k y is –

A

1p+1q−1pq\frac{1}{p} + \frac{1}{q} - \frac{1}{pq}

B

11p+1q−1pq\frac{1}{\frac{1}{p} + \frac{1}{q} - \frac{1}{pq}}

C

p + q – pq

D

p + q + pq

Answer

11p+1q−1pq\frac{1}{\frac{1}{p} + \frac{1}{q} - \frac{1}{pq}}

Explanation

Solution

p = 1 – tan2 x + tan4 x –……  = 11−(−tan⁡2x)\frac{1}{1 - ( - \tan^{2}x)}= cos2 x

q = 1 – cot2 y + cot4 y – …… = 11−(−cot⁡2y)\frac{1}{1 - ( - \cot^{2}y)}= sin2 y

Now ∑k=0∞tan⁡2kxcot⁡2ky\sum_{k = 0}^{\infty}{\tan^{2k}x\cot^{2k}y}

= 1 + tan2 x cot2 y + tan4 x cot4 y + …… 

= 11−tan⁡2xcot⁡2y\frac{1}{1 - \tan^{2}x\cot^{2}y}=11−1−cos⁡2xcos⁡2x.1−sin⁡2ysin⁡2y\frac{1}{1 - \frac{1 - \cos^{2}x}{\cos^{2}x}.\frac{1 - \sin^{2}y}{\sin^{2}y}}

= 11−(1−pp)(1−qq)\frac{1}{1 - \left( \frac{1 - p}{p} \right)\left( \frac{1 - q}{q} \right)}=pqpq−1+p+q−pq\frac{pq}{pq - 1 + p + q - pq}

= pqp+q−1\frac{pq}{p + q - 1}=11q+1p−1pq\frac{1}{\frac{1}{q} + \frac{1}{p} - \frac{1}{pq}}