Question

If $co{t}^{-1}\sqrt{cos\alpha }-ta{n}^{-1}\sqrt{cos\alpha }=x$ then sin x=

• 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 $ta{n}^2\left(\frac{\alpha }{2}\right)$

Let $co{t}^{-1}\sqrt{cos\alpha }=\theta \Rightarrow \sqrt{cos\alpha }=cot\theta \theta -ta{n}^{-1}\left(cot\theta \right)=x\Rightarrow \theta -ta{n}^{-1}\left(tan(\frac{\pi }{2}-\theta )\right)=x\Rightarrow 2\theta -\frac{\pi }{2}=x\Rightarrow sinx=-cos2\theta =-\frac{1-\frac{1}{cos\alpha }}{1+\frac{1}{cos\alpha }}=ta{n}^2\left(\frac{\alpha }{2}\right)$