prove the following identities a.((sin θ)/(1+cos 2θ))=tan θ b.((1−cos 2θ−sin θ)/(sin 2θ−cos θ))=tan θ c.((cos (x+y)+sin (x−y))/(cos 2ycos 2x))=(1/(cos (x+y)sin (y−x)))


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Question Number 55141 by 0201011563081 last updated on 18/Feb/19

$p r o v e t h e f o l l o w i n g i d e n t i t i e s$ $a . \frac{\sin θ}{1 + \cos 2 θ} = \tan θ$ $b . \frac{1 - \cos 2 θ - \sin θ}{\sin 2 θ - \cos θ} = \tan θ$ $c . \frac{\cos (x + y) + \sin (x - y)}{\cos 2 y \cos 2 x} = \frac{1}{\cos (x + y) \sin (y - x)}$
Commented by math1967 last updated on 18/Feb/19

$I t h i n k i d e n t i t y s h o u l d b e$ $\frac{\sin 2 θ}{1 + \cos 2 θ} = \frac{2 s i n θ c o s θ}{2 {c o s}^{2} θ} = t a n θ$
	Answered by math1967 last updated on 18/Feb/19

	$b) \frac{1 - c o s 2 θ - s i n θ}{s i n 2 θ - c o s θ} = \frac{2 {s i n}^{2} θ - s i n θ}{2 s i n θ c o s θ - c o s θ}$ $= \frac{s i n θ (2 s i n θ - 1)}{c o s θ (2 s i n θ - 1)} = t a n θ$
	Answered by tanmay.chaudhury50@gmail.com last updated on 18/Feb/19

	$) [c o s (x + y) + s i n (x - y)] \times [c o s (x + y) - s i n (x - y)]$ ${c o s}^{2} (x + y) - {s i n}^{2} (x - y)$ $\frac{1}{2} [2 {c o s}^{2} (x + y) - 2 {s i n}^{2} (x - y)]$ $\frac{1}{2} [1 + c o s (2 x + 2 y) - 1 + c o s (2 x - 2 y)]$ $\frac{1}{2} [2 c o s 2 x c o s 2 y]$ $= c o s 2 x c o s 2 y$ $s o \frac{c o s (x + y) + s i n (x - y)}{c o s 2 x c o s 2 y} = \frac{1}{c o s (x + y) - s i n (x - y)}$ $\frac{c o s (x + y) + s i n (x - y)}{c o s 2 x c o s 2 y} = \frac{1}{c o s (x + y) + s i n (y - x)}$ $i t h i n k t h i s i s t h e p r o b l e m . . .$ $t y p i n g e r r o r . . .$
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