# 1/sinx + 1/tanx=pi/2 Prove that 2x²+1=√5?

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Mar 8, 2018

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I hope the Question is :

If, ${\sin}^{-} 1 x + {\tan}^{-} 1 x = \frac{\pi}{2}$, prove that, $2 {x}^{2} + 1 = 5$.

Let ${\tan}^{-} 1 x = \theta \therefore \tan \theta = x \ldots \ldots \left(\star 1\right)$.

Given that, ${\sin}^{-} 1 x + {\tan}^{-} 1 x = \frac{\pi}{2} \therefore {\sin}^{-} 1 x + \theta = \frac{\pi}{2}$.

$\therefore {\sin}^{-} 1 x = \frac{\pi}{2} - \theta$.

$\therefore \sin \left(\frac{\pi}{2} - \theta\right) = x$.

$\therefore \cos \theta = x , \mathmr{and} , \sec \theta = \frac{1}{x} \ldots \ldots \left(\star 2\right)$.

$\text{But, } {\tan}^{2} \theta = {\sec}^{2} \theta - 1$.

$\therefore {x}^{2} = \frac{1}{x} ^ 2 - 1 , \mathmr{and} , {x}^{4} + {x}^{2} - 1 = 0$.

This is a quadr. eqn. in ${x}^{2}$.

Solving it with the help of the quadr. formula, we get,

${x}^{2} = \frac{- 1 \pm \sqrt{1 + 4}}{2} = \frac{- 1 \pm \sqrt{5}}{2}$.

Since, ${x}^{2} > 0 , {x}^{2} \ne \frac{- 1 - \sqrt{5}}{2}$.

$\therefore {x}^{2} = \frac{- 1 + \sqrt{5}}{2}$, or what is the same as to say that,

$2 {x}^{2} + 1 = \sqrt{5}$.

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