If tan theta = t, then tan 2theta + sec 2thet | vedclass

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(A) Given $	an 	heta = t.$We know that $	an 2	heta = \frac{2	an 	heta}{1 - 	an^2 	heta} = \frac{2t}{1 - t^2}$ and $\sec 2	heta = \frac{1}{\co

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$\frac{1 + t}{1 - t}$

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(A) Given $	an 	heta = t.$We know that $	an 2	heta = \frac{2	an 	heta}{1 - 	an^2 	heta} = \frac{2t}{1 - t^2}$ and $\sec 2	heta = \frac{1}{\cos 2	heta} = \frac{1 + 	an^2 	heta}{1 - 	an^2 	heta} = \frac{1 + t^2}{1 - t^2}.$Therefore,$	an 2	heta + \sec 2	heta = \frac{2t}{1 - t^2} + \frac{1 + t^2}{1 - t^2}.$$= \frac{2t + 1 + t^2}{1 - t^2} = \frac{(1 + t)^2}{(1 - t)(1 + t)}.$$= \frac{1 + t}{1 - t}.$

If $\tan \theta = t,$ then $\tan 2\theta + \sec 2\theta = $

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