Question

$co{t}^{-1}\left[\right(cos\alpha {)}^{\frac{1}{2}}]-ta{n}^{-1}[\left(cos\alpha {)}^{\frac{1}{2}}\right]=x$, then $sinx=$

• A. $ta{n}^2\left(\frac{\alpha }{2}\right)$
• B. $co{t}^2\left(\frac{\alpha }{2}\right)$
• C. $tan\alpha$
• D. $cot\left(\frac{\alpha }{2}\right)$

Answer 1

The correct option is A $ta{n}^2\left(\frac{\alpha }{2}\right)$

$ta{n}^{-1}\left[\frac{1}{\sqrt{cos\alpha }}\right]-ta{n}^{-1}\left[\sqrt{cos\alpha }\right]=x$
$\Rightarrow ta{n}^{-1}\left[\frac{\frac{1}{\sqrt{cos\alpha }}-\sqrt{cos\alpha }}{1+\frac{\sqrt{cos\alpha }}{\sqrt{cos\alpha }}}\right]=x\Rightarrow tanx=\frac{1-cos\alpha }{2\sqrt{cos\alpha }}$
$\therefore sinx=\frac{1-cos\alpha }{1+cos\alpha }=\frac{2si{n}^2\frac{\alpha }{2}}{2co{s}^2\frac{\alpha }{2}}=ta{n}^2\left(\frac{\alpha }{2}\right)$.