show that a) ((1−cos2A)/(1+cos2A)) ≡ tan^2 A b) ((sin2A)/(1+cos2A))≡ tanA.


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Question Number 41394 by Rio Michael last updated on 07/Aug/18

$s h o w t h a t a) \frac{1 - c o s 2 A}{1 + c o s 2 A} \equiv {t a n}^{2} A$ $b) \frac{s i n 2 A}{1 + c o s 2 A} \equiv t a n A .$
Commented by mondodotto@gmail.com last updated on 07/Aug/18

$in part B it should be proved$ $to tanA not \tan 2 A$
	Answered by alex041103 last updated on 06/Aug/18

	$a) c o s (2 A) = {c o s}^{2} A - {s i n}^{2} A$ $a n d {c o s}^{2} A + {s i n}^{2} A = 1$ $\Rightarrow c o s (2 A) = 2 {c o s}^{2} A - 1 =$ $= 1 - 2 {s i n}^{2} A$ $\Rightarrow 1 - c o s 2 A = 2 {s i n}^{2} A$ $1 + c o s 2 A = 2 {c o s}^{2} A$ $\Rightarrow \frac{1 - c o s 2 A}{1 + c o s 2 A} = \frac{2 {s i n}^{2} A}{2 {c o s}^{2} A} = {(\frac{s i n A}{c o s A})}^{2} = {t a n}^{2} A$ $b) s i n (2 A) = 2 s i n A c o s A$ $\Rightarrow \frac{s i n 2 A}{1 + c o s 2 A} = \frac{2 s i n A c o s A}{2 {c o s}^{2} A} = \frac{s i n A}{c o s A} = t a n A$ $W e p r o v e d t h a t :$ $a) \frac{1 - c o s 2 A}{1 + c o s 2 A} = {t a n}^{2} A$ $b) \frac{s i n 2 A}{1 + c o s 2 A} = t a n A$
	Commented by alex041103 last updated on 06/Aug/18

	$I s h o u l d p o i n t o u t t h a t {t h e r e}^{'} s a m i s t a k e i n t h e p r o b l e m .$
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