If-tan-x-2-csc-x-sin-x-then-tan-2-x-2-

Question Number 189468 by cortano12 last updated on 17/Mar/23

If \tan (\frac{x}{2}) = \csc x - \sin x, then

\tan^{2} (\frac{x}{2}) = ?

Commented by mehdee42 last updated on 17/Mar/23

t h e g i v e n r e l a t i o n i s w r o n g .

t a n (\frac{x}{2}) = c s c x - c o t x

Commented by cortano12 last updated on 17/Mar/23

why wrong ?

Commented by Frix last updated on 17/Mar/23

Why do you think {it}^{'} s a relation ?

Commented by mehdee42 last updated on 17/Mar/23

s o r r y . i t w a s m y m i s t a k e . t h i s i s a n e q u a t i o n .

Answered by horsebrand11 last updated on 17/Mar/23

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

l e t \tan (\frac{x}{2}) = p \Rightarrow p^{2} = \tan^{2} (\frac{x}{2})

\Rightarrow p = \frac{1 + p^{2}}{2 p} - \frac{2 p}{1 + p^{2}}

\Rightarrow p = \frac{{(1 + p^{2})}^{2} - {(2 p)}^{2}}{2 p (1 + p^{2})}

\Rightarrow 2 p^{2} (1 + p^{2}) = p^{4} - 2 p^{2} + 1

l e t k = p^{2}, k > 0

\Rightarrow 2 k (1 + k) = k^{2} - 2 k + 1

\Rightarrow 2 k + 2 k^{2} = k^{2} - 2 k + 1

\Rightarrow k^{2} + 4 k - 1 = 0

\Rightarrow k = p^{2} = \frac{- 4 + \sqrt{20}}{2} = \sqrt{5} - 2

Answered by Frix last updated on 17/Mar/23

\tan \frac{x}{2} = \csc x - \sin x

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

t = \frac{{(t^{2} - 1)}^{2}}{2 t (t^{2} + 1)}

t^{4} + 4 t^{2} - 1 = 0

t^{2} = - 2 + \sqrt{5} (for t \in R)

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