prove-that-2-tan-1-2-3-sin-1-12-13-

Question Number 114108 by bemath last updated on 17/Sep/20

p r o v e t h a t 2 \tan^{- 1} (\frac{2}{3}) = \sin^{- 1} (\frac{12}{13})

Answered by bobhans last updated on 17/Sep/20

l e t \tan^{- 1} (\frac{2}{3}) = x \to \tan x = \frac{2}{3}

a n d {\begin{cases} \sin x = \frac{2}{\sqrt{13}} \\ \cos x = \frac{3}{\sqrt{13}} \end{cases}

\Leftrightarrow 2 x = \sin^{- 1} (\frac{12}{13})

\Leftrightarrow \sin (2 x) = \sin (\sin^{- 1} (\frac{12}{13}))

\Rightarrow 2 \sin x \cos x = \frac{12}{13}

\Rightarrow 2 (\frac{2}{\sqrt{13}}) (\frac{3}{\sqrt{13}}) = \frac{12}{13}

Answered by Dwaipayan Shikari last updated on 17/Sep/20

{t a n}^{- 1} (\frac{\frac{2}{3} + \frac{2}{3}}{1 - \frac{4}{9}}) = {t a n}^{- 1} \frac{12}{5}

t a n θ = \frac{12}{5}

s e c θ = \frac{13}{5}

c o s θ = \frac{5}{13} a n d s i n θ = \sqrt{1 - \frac{5^{2}}{13^{2}}} = \frac{12}{13}

θ = {s i n}^{- 1} \frac{12}{13}

θ = 2 {t a n}^{- 1} \frac{2}{3} (W h i c h i s t r u e)

Answered by physicstutes last updated on 17/Sep/20

if \sin θ = \frac{12}{13}, then \tan θ = \frac{12}{5}

\tan (\frac{θ}{2}) = \frac{\sin θ}{1 + \cos θ}

\cos θ = \frac{5}{13} \Rightarrow \tan (\frac{θ}{2}) = \frac{\frac{12}{13}}{1 + \frac{5}{13}} = \frac{12}{18} = \frac{2}{3}

hence \tan (\frac{θ}{2}) = \frac{2}{3} \Rightarrow θ = 2 \tan^{- 1} (\frac{2}{3})

thus, θ = \sin^{- 1} (\frac{12}{13}) = 2 \tan^{- 1} (\frac{2}{3})