If Cosθ=((x cosβ − y)/(x − y cosβ)) then prove that, tan(θ/2) =(√(((x−y)/(x+y)) )) tan(β/2)


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Question Number 71918 by lalitchand last updated on 22/Oct/19

$If Cos θ = \frac{x \cos β - y}{x - y \cos β} then prove that, \tan \frac{θ}{2} = \sqrt{\frac{x - y}{x + y}} \tan \frac{β}{2}$
Commented by Prithwish sen last updated on 22/Oct/19

$Using componendo dividendo$ $\frac{1 - \cos θ}{1 + \cos θ} = \frac{(1 - \cos β) (x + y)}{(1 + \cos β) (x - y)}$ $\tan \frac{θ}{2} = \sqrt{\frac{x + y}{x - y}} \tan \frac{β}{2}$ $please check the question .$
Commented by lalitchand last updated on 22/Oct/19

$Is there mistake sir ? ?$
	Answered by mind is power last updated on 22/Oct/19
	$cos(β)=2cos^2 ((β/2))−1=((1−tg^2 ((β/2)))/(1+tg^2 ((β/2)))) assum x≥y x+y#0 β,θ∈[0,π] ((xcos(β)−y)/(x−ycos(β)))=((x(1−tg^2 (β))−y(1+tg^2 ((β/2)))/(x(1+tg^2 (β))−y(1−tg^2 (β))))=(((x−y)−(x+y)tg^2 ((β/2)))/((x−y)+(x+y)tg^2 ((β/2)))) cos(θ)=((1−tg^2 ((θ/2)))/(1+tg^2 ((θ/2))))=(((x−y)−(x+y)tg^2 ((β/2)))/((x−y)+(x+y)tg^2 ((β/2))))=((1−((x+y)/(x−y))tg^2 ((β/2)))/(1+((x+y)/(x−y))tg^2 ((β/2)))) let f(x)=((1−x)/(1+x)) x∈R−{−1}⇒f′(x)=((−2)/((1+x)^2 ))<0 decreasing...I ⇒f(x)=f(y),x=y cos(θ)=((1−tg^2 ((θ/2)))/(1+tg^2 ((θ/2))))=(((x−y)−(x+y)tg^2 ((β/2)))/((x−y)+(x+y)tg^2 ((β/2))))=((1−((x+y)/(x−y))tg^2 ((β/2)))/(1+((x+y)/(x−y))tg^2 ((β/2)))) ⇔f(tg^2 ((θ/2)))=f(((x+y)/(x−y))tg^2 ((β/2)))⇒by I tg^2 ((θ/2))=((x+y)/(x−y))tg^2 ((β/2))⇒tg((θ/2))=(√((x+y)/(x−y)))tg((β/2))$
	$\cos (β) = {2 \cos}^{2} (\frac{β}{2}) - 1 = \frac{1 - {tg}^{2} (\frac{β}{2})}{1 + {tg}^{2} (\frac{β}{2})}$ $You can't use 'macro parameter character #' in math mode$ $\frac{xcos (β) - y}{x - ycos (β)} = \frac{x (1 - {tg}^{2} (β)) - y (1 + {tg}^{2} (\frac{β}{2})}{x (1 + {tg}^{2} (β)) - y (1 - {tg}^{2} (β))} = \frac{(x - y) - (x + y) {tg}^{2} (\frac{β}{2})}{(x - y) + (x + y) {tg}^{2} (\frac{β}{2})}$ $\cos (θ) = \frac{1 - {tg}^{2} (\frac{θ}{2})}{1 + {tg}^{2} (\frac{θ}{2})} = \frac{(x - y) - (x + y) {tg}^{2} (\frac{β}{2})}{(x - y) + (x + y) {tg}^{2} (\frac{β}{2})} = \frac{1 - \frac{x + y}{x - y} {tg}^{2} (\frac{β}{2})}{1 + \frac{x + y}{x - y} {tg}^{2} (\frac{β}{2})}$ $let f (x) = \frac{1 - x}{1 + x} x \in R - {- 1} \Rightarrow f^{'} (x) = \frac{- 2}{{(1 + x)}^{2}} < 0 decreasing . . . I$ $\Rightarrow f (x) = f (y), x = y$ $\cos (θ) = \frac{1 - {tg}^{2} (\frac{θ}{2})}{1 + {tg}^{2} (\frac{θ}{2})} = \frac{(x - y) - (x + y) {tg}^{2} (\frac{β}{2})}{(x - y) + (x + y) {tg}^{2} (\frac{β}{2})} = \frac{1 - \frac{x + y}{x - y} {tg}^{2} (\frac{β}{2})}{1 + \frac{x + y}{x - y} {tg}^{2} (\frac{β}{2})}$ $\Leftrightarrow f ({tg}^{2} (\frac{θ}{2})) = f (\frac{x + y}{x - y} {tg}^{2} (\frac{β}{2})) \Rightarrow by I$ ${tg}^{2} (\frac{θ}{2}) = \frac{x + y}{x - y} {tg}^{2} (\frac{β}{2}) \Rightarrow tg (\frac{θ}{2}) = \sqrt{\frac{x + y}{x - y}} tg (\frac{β}{2})$
	Commented by lalitchand last updated on 22/Oct/19

	$I dont understand sir ? if u dont mind tell me easy way$
	Commented by mind is power last updated on 23/Oct/19

	$i will try later other solution$
	Answered by Tanmay chaudhury last updated on 22/Oct/19

	$x c o s θ - y c o s β c o s θ = x c o s β - y$ $x (c o s θ - c o s β) = y (c o s β c o s θ - 1)$ $\frac{x}{y} = \frac{c o s β c o s θ - 1}{c o s θ - c o s β}$ $\frac{x - y}{x + y} = \frac{c o s β c o s θ - 1 - c o s θ + c o s β}{c o s β c o s θ - 1 + c o s θ - c o s β}$ $= \frac{c o s θ (c o s β - 1) + 1 (c o s β - 1)}{c o s θ (1 + c o s β) - 1 (1 + c o s β)}$ $= \frac{(1 + c o s θ) (1 - c o s β) \times - 1}{(1 + c o s β) (1 - c o s θ) \times - 1}$ $= \frac{2 {c o s}^{2} \frac{θ}{2} \times 2 {s i n}^{2} \frac{β}{2}}{2 {c o s}^{2} \frac{β}{2} \times 2 {s i n}^{2} \frac{θ}{2}}$ $= \frac{{t a n}^{2} \frac{β}{2}}{{t a n}^{2} \frac{θ}{2}}$ $t a n \frac{β}{2} = \sqrt{\frac{x - y}{x + y}} \times t a n \frac{θ}{2}$ $p l s c h e c k . . .$
	Commented by mind is power last updated on 23/Oct/19

	$nice sir creative solution$
	Commented by Tanmay chaudhury last updated on 23/Oct/19

	$t h a n k y o u s i r$
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