If tan left( frac{pi}{4} + theta right) + tan | vedclass

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(D) We know that $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Using these,$

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$2$

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(D) We know that $	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$ and $	an(A - B) = \frac{	an A - 	an B}{1 + 	an A 	an B}$.Using these,$	an \left( \frac{\pi}{4} + 	heta ight) = \frac{1 + 	an 	heta}{1 - 	an 	heta}$ and $	an \left( \frac{\pi}{4} - 	heta ight) = \frac{1 - 	an 	heta}{1 + 	an 	heta}$.Adding these two expressions:$\frac{1 + 	an 	heta}{1 - 	an 	heta} + \frac{1 - 	an 	heta}{1 + 	an 	heta} = \frac{(1 + 	an 	heta)^2 + (1 - 	an 	heta)^2}{(1 - 	an 	heta)(1 + 	an 	heta)}$$= \frac{1 + 	an^2 	heta + 2 	an 	heta + 1 + 	an^2 	heta - 2 	an 	heta}{1 - 	an^2 	heta}$$= \frac{2(1 + 	an^2 	heta)}{1 - 	an^2 	heta}$Since $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$,we have $\sec 2	heta = \frac{1 + 	an^2 	heta}{1 - 	an^2 	heta}$.Thus,the expression becomes $2 \sec 2	heta$.Comparing this with $\lambda \sec 2	heta$,we get $\lambda = 2$.

If $\tan \left( \frac{\pi}{4} + \theta \right) + \tan \left( \frac{\pi}{4} - \theta \right) = \lambda \sec 2\theta$,then $\lambda$ =

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