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(D) અનંત ગુણોત્તર શ્રેણી આપેલ છે:$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

Vedclass · Accepted Answer

$xyz = x + y + z$ અને $xyz = xy + z$

Vedclass · Answer

(D) અનંત ગુણોત્તર શ્રેણી આપેલ છે:$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}$હવે,$xy$ ની ગણતરી કરીએ:$xy = \frac{1}{\sin^2\phi} \cdot \frac{1}{\cos^2\phi} = \frac{1}{\sin^2\phi \cos^2\phi}$વિકલ્પ $(b)$ ચકાસો:$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)}$આમ,$xyz = xy + z$ સાચું છે.વિકલ્પ $(c)$ ચકાસો:$x + y + z = \frac{1}{\sin^2\phi} + \frac{1}{\cos^2\phi} + z = \frac{\cos^2\phi + \sin^2\phi}{\sin^2\phi \cos^2\phi} + z = \frac{1}{\sin^2\phi \cos^2\phi} + z = xy + z = xyz$.તેથી,$(b)$ અને $(c)$ બંને સાચા હોવાથી,વિકલ્પ $(d)$ સાચો જવાબ છે.

$0 < \phi < \frac{\pi}{2}$ માટે,જો $x = \sum_{n=0}^\infty \cos^{2n}\phi$,$y = \sum_{n=0}^\infty \sin^{2n}\phi$,અને $z = \sum_{n=0}^\infty \cos^{2n}\phi \sin^{2n}\phi$ હોય,તો:

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