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

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Question

(A) We use the expansion formulas $	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}$.

Vedclass · Accepted Answer

$2	an 2	heta $

Vedclass · Answer

(A) We use the expansion formulas $	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}$.Given expression: $	an \left( \frac{\pi }{4} + 	heta ight) - 	an \left( \frac{\pi }{4} - 	heta ight)$.Substituting $	an \frac{\pi }{4} = 1$,we get:$= \frac{1 + 	an 	heta}{1 - 	an 	heta} - \frac{1 - 	an 	heta}{1 + 	an 	heta}$.Taking the common denominator:$= \frac{(1 + 	an 	heta)^2 - (1 - 	an 	heta)^2}{(1 - 	an 	heta)(1 + 	an 	heta)}$.Using $(a+b)^2 - (a-b)^2 = 4ab$ and $(a-b)(a+b) = a^2 - b^2$:$= \frac{4 	an 	heta}{1 - 	an^2 	heta}$.$= 2 \left( \frac{2 	an 	heta}{1 - 	an^2 	heta} ight)$.Since $	an 2	heta = \frac{2 	an 	heta}{1 - 	an^2 	heta}$,the expression simplifies to:$= 2 	an 2	heta$.

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

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