A) \[\tan \frac{\beta }{2}\tan {{15}^{o}}\]
B) \[\tan \frac{\beta }{2}\]
C) \[\tan {{15}^{o}}\]
D) None of these
Correct Answer: A
Solution :
Given \[\cos x-\frac{\cot \beta \sin x}{2}=\frac{\sqrt{3}}{2}\] \[\Rightarrow \]\[\frac{1-{{\tan }^{2}}\frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}-\frac{\cot \beta \tan \frac{x}{2}}{1+{{\tan }^{2}}\frac{x}{2}}=\frac{\sqrt{3}}{2}\] \[\Rightarrow \]\[1-{{\tan }^{2}}\frac{x}{2}-\cot \beta \tan \frac{x}{2}\] \[=\frac{\sqrt{3}}{2}\left( 1+{{\tan }^{2}}\frac{x}{2} \right)\] \[\Rightarrow \]\[(2+\sqrt{3}){{\tan }^{2}}\frac{x}{2}+2\cot \beta \tan \frac{x}{2}\] \[+(\sqrt{3}-2)=0\] \[\Rightarrow \]\[\tan \frac{x}{2}=\frac{-2\cot \beta \pm \sqrt{4{{\cot }^{2}}\beta +4}}{2(2+\sqrt{3})}\] \[\Rightarrow \]\[\tan \frac{x}{2}=\frac{-\cot \beta +\cos ec\beta }{2+\sqrt{3}}\] \[\Rightarrow \]\[\tan \frac{x}{2}=\frac{-\cot \beta -\cos ec\beta }{2+\sqrt{3}}\] \[\Rightarrow \]\[\tan \frac{x}{2}=\tan \frac{\beta }{2}\tan {{15}^{o}}\]
You need to login to perform this action.
You will be redirected in
3 sec
