If sec theta = x + frac{1}{4x} where 0^{circ} | vedclass

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(B) Given $\sec 	heta = x + \frac{1}{4x} = \frac{4x^2 + 1}{4x}$.We know the identity $	an^2 	heta = \sec^2 	heta - 1$.$	an 	heta = \sqrt{\sec^2

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$2x$

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(B) Given $\sec 	heta = x + \frac{1}{4x} = \frac{4x^2 + 1}{4x}$.We know the identity $	an^2 	heta = \sec^2 	heta - 1$.$	an 	heta = \sqrt{\sec^2 	heta - 1} = \sqrt{\left(x + \frac{1}{4x}ight)^2 - 1}$.$	an 	heta = \sqrt{x^2 + 2(x)(\frac{1}{4x}) + \frac{1}{16x^2} - 1} = \sqrt{x^2 + \frac{1}{2} + \frac{1}{16x^2} - 1} = \sqrt{x^2 - \frac{1}{2} + \frac{1}{16x^2}}$.This can be written as $\sqrt{\left(x - \frac{1}{4x}ight)^2} = x - \frac{1}{4x}$ (since $0^{\circ}  0$).Now,$\sec 	heta + 	an 	heta = \left(x + \frac{1}{4x}ight) + \left(x - \frac{1}{4x}ight) = 2x$.

If $\sec \theta = x + \frac{1}{4x}$ where $0^{\circ} < \theta < 90^{\circ}$,then $\sec \theta + \tan \theta$ is equal to:

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