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

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Question

(A) We use the expansion formulas for $	an(A+B)$ and $	an(A-B)$:$	an \left( \frac{\pi }{4} + 	heta ight) = \frac{	an(\pi/4) + 	an 	heta}{1 -

Vedclass · Accepted Answer

$2	an 2	heta $

Vedclass · Answer

(A) We use the expansion formulas for $	an(A+B)$ and $	an(A-B)$:$	an \left( \frac{\pi }{4} + 	heta ight) = \frac{	an(\pi/4) + 	an 	heta}{1 - 	an(\pi/4)	an 	heta} = \frac{1 + 	an 	heta}{1 - 	an 	heta}$$	an \left( \frac{\pi }{4} - 	heta ight) = \frac{	an(\pi/4) - 	an 	heta}{1 + 	an(\pi/4)	an 	heta} = \frac{1 - 	an 	heta}{1 + 	an 	heta}$Now,subtract the two expressions:$\frac{1 + 	an 	heta}{1 - 	an 	heta} - \frac{1 - 	an 	heta}{1 + 	an 	heta} = \frac{(1 + 	an 	heta)^2 - (1 - 	an 	heta)^2}{(1 - 	an 	heta)(1 + 	an 	heta)}$$= \frac{(1 + 2	an 	heta + 	an^2 	heta) - (1 - 2	an 	heta + 	an^2 	heta)}{1 - 	an^2 	heta}$$= \frac{4	an 	heta}{1 - 	an^2 	heta}$$= 2 \left( \frac{2	an 	heta}{1 - 	an^2 	heta} ight)$Since $	an 2	heta = \frac{2	an 	heta}{1 - 	an^2 	heta}$,the expression becomes $2	an 2	heta$.

$\tan \left( \frac{\pi }{4} + \theta \right) - \tan \left( \frac{\pi }{4} - \theta \right) = $

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