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Question Number 12572 by tawa last updated on 25/Apr/17

prove that if,

\sin (θ) = \frac{1 - x}{1 + x} then \tan (\frac{x}{4} - \frac{θ}{2}) = \sqrt{x}

Commented by mrW1 last updated on 26/Apr/17

I t h i n k y o u m e a n t a n (\frac{π}{4} - \frac{θ}{2}) = \sqrt{x}

Answered by mrW1 last updated on 26/Apr/17

\sin θ = \frac{2 \tan \frac{θ}{2}}{1 + \tan^{2} \frac{θ}{2}} = \frac{1 - x}{1 + x}

l e t t = \tan \frac{θ}{2}

\Rightarrow \frac{2 t}{1 + t^{2}} = \frac{1 - x}{1 + x}

(1 - x) t^{2} - 2 (1 + x) t + (1 - x) = 0

t = \frac{2 (1 + x) \pm \sqrt{4 {(1 + x)}^{2} - 4 {(1 - x)}^{2}}}{2 (1 - x)}

= \frac{(1 + x) \pm 2 \sqrt{x}}{(1 - x)} = \frac{{(1 \pm \sqrt{x})}^{2}}{(1 + \sqrt{x}) (1 - \sqrt{x})}

\Rightarrow t = \frac{1 + \sqrt{x}}{1 - \sqrt{x}} o r \frac{1 - \sqrt{x}}{1 + \sqrt{x}}

\tan (\frac{π}{4} - \frac{θ}{2}) = \frac{\tan \frac{π}{4} - \tan \frac{θ}{2}}{1 + \tan \frac{π}{4} \times \tan \frac{θ}{2}}

= \frac{1 - t}{1 + t} = \frac{1 - \frac{1 - \sqrt{x}}{1 + \sqrt{x}}}{1 + \frac{1 - \sqrt{x}}{1 + \sqrt{x}}} = \sqrt{x}

o r

= \frac{1 - t}{1 + t} = \frac{1 - \frac{1 + \sqrt{x}}{1 - \sqrt{x}}}{1 + \frac{1 + \sqrt{x}}{1 - \sqrt{x}}} = - \sqrt{x}

Commented by tawa last updated on 26/Apr/17

God bless you sir . i really appreciate your effort

Answered by ajfour last updated on 26/Apr/17

\tan (\frac{π}{4} - \frac{θ}{2}) = \frac{1 - \tan (θ / 2)}{1 + \tan (θ / 2)}

\sin θ = \frac{2 \tan (θ / 2)}{1 + \tan^{2} (θ / 2)} = \frac{1 - x}{1 + x} = \frac{p}{q} (s a y)

\frac{q - p}{q + p} = {[\frac{1 - \tan (θ / 2)}{1 + \tan (θ / 2)}]}^{2} = \frac{(1 + x) - (1 - x)}{(1 + x) + (1 - x)}

\Rightarrow \tan^{2} (\frac{π}{4} - \frac{θ}{2}) = \frac{2 x}{2} = x

\Rightarrow \tan (\frac{π}{4} - \frac{θ}{2}) = \pm \sqrt{x} .

Commented by mrW1 last updated on 26/Apr/17

C a n y o u p l e a s e c h e c k :

I n m y w a y I g e t \tan (\frac{π}{4} - \frac{θ}{2}) = \pm \sqrt{x}

b u t i n y o u r w a y y o u g e t o n l y \sqrt{x .}

F o r e x a m p l e :

x = 3

\sin θ = \frac{1 - 3}{1 + 3} = - \frac{2}{4} = - \frac{1}{2}

\Rightarrow θ = \frac{7 π}{6} o r - \frac{π}{6}

\frac{π}{4} - \frac{θ}{2} = \frac{π}{4} - \frac{7 π}{12} = - \frac{π}{3} o r = \frac{π}{4} + \frac{π}{12} = \frac{π}{3}

\tan (\frac{π}{4} - \frac{θ}{2}) = \tan (- \frac{π}{3}) = - \sqrt{3}

o r

\tan (\frac{π}{4} - \frac{θ}{2}) = \tan (\frac{π}{3}) = \sqrt{3}

I s s o m e t h i n g w r o n g h e r e ?

Commented by ajfour last updated on 26/Apr/17

i h a d a s s u m e d \sin θ = \frac{p}{q} > 0,

{s h o u l d n}^{'} t h a v e .

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