If x = sec theta + tan theta , then x + frac{ | vedclass

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(B) Given that $x = \sec 	heta + 	an 	heta$.We know the identity $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec 	heta - 	an \

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$2 \sec 	heta$

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(B) Given that $x = \sec 	heta + 	an 	heta$.We know the identity $\sec^2 	heta - 	an^2 	heta = 1$,which can be written as $(\sec 	heta - 	an 	heta)(\sec 	heta + 	an 	heta) = 1$.Therefore,$\frac{1}{x} = \frac{1}{\sec 	heta + 	an 	heta} = \sec 	heta - 	an 	heta$.Now,calculating $x + \frac{1}{x}$:$x + \frac{1}{x} = (\sec 	heta + 	an 	heta) + (\sec 	heta - 	an 	heta)$.$x + \frac{1}{x} = 2 \sec 	heta$.

If $x = \sec \theta + \tan \theta ,$ then $x + \frac{1}{x} = $

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