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

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(D) We know that $	an \left( \frac{\pi }{4} + 	heta ight) = \frac{1 + 	an 	heta}{1 - 	an 	heta}$ and $	an \left( \frac{\pi }{4} - 	heta ig

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$2$

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(D) We know that $	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,we get $\frac{1 + 	an 	heta}{1 - 	an 	heta} + \frac{1 - 	an 	heta}{1 + 	an 	heta} = \frac{(1 + 	an 	heta)^2 + (1 - 	an 	heta)^2}{(1 - 	an 	heta)(1 + 	an 	heta)}$.$= \frac{1 + 	an^2 	heta + 2 	an 	heta + 1 + 	an^2 	heta - 2 	an 	heta}{1 - 	an^2 	heta} = \frac{2(1 + 	an^2 	heta)}{1 - 	an^2 	heta}$.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 $2 \sec 2	heta = \lambda \sec 2	heta$.Therefore,$\lambda = 2$.

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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