# cos⁻¹(sqrtcos α)−tan⁻¹(sqrtcos α)=x ,then what is the value of sin x ?

## (a) tan^2(α/2) (b) cot^2(α/2) (c) tanα (d) cot (α/2)

May 15, 2018

$\sin x = \tan \left(\frac{\alpha}{2}\right) - \cos \frac{\alpha}{\sqrt{2} \cos \left(\frac{\alpha}{2}\right)}$

#### Explanation:

Let $\sqrt{\cos} \alpha = m$

$\rightarrow {\cos}^{- 1} \left(m\right) - {\tan}^{- 1} \left(m\right) = x$

Let ${\cos}^{- 1} m = y$ then $\cos y = m$

$\rightarrow \sin y = \sqrt{1 - {\cos}^{2} y} = \sqrt{1 - {m}^{2}}$

$\rightarrow y = {\sin}^{- 1} \left(\sqrt{1 - {m}^{2}}\right) = {\cos}^{- 1} m$

Also, let ${\tan}^{- 1} m = z$ then $\tan z = m$

$\rightarrow \sin z = \frac{1}{\csc} z = \frac{1}{\sqrt{1 + {\cot}^{2} z}} = \frac{1}{\sqrt{1 + {\left(\frac{1}{m}\right)}^{2}}} = \frac{m}{\sqrt{1 + {m}^{2}}}$

$\rightarrow z = {\sin}^{- 1} \left(\frac{m}{\sqrt{1 + {m}^{2}}}\right) = {\tan}^{- 1} m$

$\rightarrow {\cos}^{- 1} \left(m\right) - {\tan}^{- 1} \left(m\right)$

$= {\sin}^{- 1} \left(\sqrt{1 - {m}^{2}}\right) - {\sin}^{- 1} \left(\frac{m}{\sqrt{1 + {m}^{2}}}\right)$

$= {\sin}^{-} 1 \left(\sqrt{1 - {m}^{2}} \cdot \sqrt{1 - {\left(\frac{m}{\sqrt{1 + {m}^{2}}}\right)}^{2}} - \left(\frac{m}{\sqrt{1 + {m}^{2}}}\right) \cdot \sqrt{1 - {\left(\sqrt{1 - {m}^{2}}\right)}^{2}}\right)$

$= {\sin}^{- 1} \left(\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} - \cos \frac{\alpha}{\sqrt{1 + \cos \alpha}}\right)$

$= {\sin}^{- 1} \left(\tan \left(\frac{\alpha}{2}\right) - \cos \frac{\alpha}{\sqrt{2} \cos \left(\frac{\alpha}{2}\right)}\right) = x$

$\rightarrow \sin x = \sin \left({\sin}^{- 1} \left(\tan \left(\frac{\alpha}{2}\right) - \cos \frac{\alpha}{\sqrt{2} \cos \left(\frac{\alpha}{2}\right)}\right)\right) = \tan \left(\frac{\alpha}{2}\right) - \cos \frac{\alpha}{\sqrt{2} \cos \left(\frac{\alpha}{2}\right)}$