For 0

Question

(D) Given the infinite geometric series:$x = 1 + \cos^2\phi + \cos^4\phi + \dots = \frac{1}{1 - \cos^2\phi} = \frac{1}{\sin^2\phi}$$y = 1 + \sin^2\phi

Vedclass · Accepted Answer

$xyz = x + y + z$ and $xyz = xy + z$

Vedclass · Answer

(D) Given the infinite geometric series:$x = 1 + \cos^2\phi + \cos^4\phi + \dots = \frac{1}{1 - \cos^2\phi} = \frac{1}{\sin^2\phi}$$y = 1 + \sin^2\phi + \sin^4\phi + \dots = \frac{1}{1 - \sin^2\phi} = \frac{1}{\cos^2\phi}$$z = 1 + \cos^2\phi \sin^2\phi + \cos^4\phi \sin^4\phi + \dots = \frac{1}{1 - \cos^2\phi \sin^2\phi}$Now,calculate $xy$:$xy = \frac{1}{\sin^2\phi} \cdot \frac{1}{\cos^2\phi} = \frac{1}{\sin^2\phi \cos^2\phi}$Check option $(b)$:$xy + z = \frac{1}{\sin^2\phi \cos^2\phi} + \frac{1}{1 - \cos^2\phi \sin^2\phi} = \frac{1 - \cos^2\phi \sin^2\phi + \sin^2\phi \cos^2\phi}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)} = \frac{1}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)}$$xyz = \left(\frac{1}{\sin^2\phi \cos^2\phi}\right) \left(\frac{1}{1 - \cos^2\phi \sin^2\phi}\right) = \frac{1}{\sin^2\phi \cos^2\phi (1 - \cos^2\phi \sin^2\phi)}$Thus,$xyz = xy + z$ is correct.Check option $(c)$:$x + y + z = \frac{1}{\sin^2\phi} + \frac{1}{\cos^2\phi} + z = \frac{\cos^2\phi + \sin^2\phi}{\sin^2\phi \cos^2\phi} + z = \frac{1}{\sin^2\phi \cos^2\phi} + z = xy + z = xyz$.Since both $(b)$ and $(c)$ are correct,option $(d)$ is the correct choice.

For $0 < \phi < \frac{\pi}{2}$,if $x = \sum_{n=0}^\infty \cos^{2n}\phi$,$y = \sum_{n=0}^\infty \sin^{2n}\phi$,and $z = \sum_{n=0}^\infty \cos^{2n}\phi \sin^{2n}\phi$,then:

Similar Questions

The radius of the circle whose arc of length $15 \ cm$ makes an angle of $3/4$ radian at the centre is ..... $cm$.

$\frac{{\sin \theta }}{{1 - \cot \theta }} + \frac{{\cos \theta }}{{1 - \tan \theta }} = $

Let $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ where $x \in R$ and $k \ge 1$. Then $f_4(x) - f_6(x)$ is equal to:

Which of the following is correct?

If $90^\circ < A < 180^\circ$ and $\sin A = \frac{4}{5}$,then $\tan \frac{A}{2}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module