Question

$\frac{(co{t}^{2}15°-1)}{(co{t}^{2}15°+1)}=$

• A. $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• B. $\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$
• C. $\raisebox{1ex}{$3\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$4$}\right.$
• D. $\sqrt{3}$

Answer 1

The correct option is B

$\raisebox{1ex}{$\sqrt{3}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$$\begin{gathered} \mathbf{Explanation}\mathbf{}\mathbf{for}\mathbf{}\mathbf{the}\mathbf{}\mathbf{correct}\mathbf{}\mathbf{option}\mathbf{:} \\ \mathbf{option}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{:} \\ \mathrm{Given}, \end{gathered}$$

$\frac{(co{t}^{2}15°-1)}{(co{t}^{2}15°+1)}=$

We know that,

$$\begin{gathered} \Rightarrow cot\theta =\frac{1}{\tan \left(\theta \right)} \\ =\frac{{\left(\frac{1}{\tan {15}^{°}}\right)}^2-1}{{\left(\frac{1}{\tan {15}^{°}}\right)}^2+1} \\ =\frac{{\displaystyle \frac{1-{\tan }^2{15}^{°}}{{\tan }^2{15}^{°}}}}{{\displaystyle \frac{1+{\tan }^2{15}^{°}}{{\tan }^2{15}^{°}}}} \\ =\frac{{\displaystyle 1-{\tan }^2{15}^{°}}}{{\displaystyle 1+{\tan }^2{15}^{°}}} \\ \Rightarrow \cos \left(2\theta \right)=\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta } \\ \Rightarrow \theta ={15}^{°} \\ \Rightarrow \cos \left(2\times {15}^{°}\right)=\cos \left({30}^{{°}^{}}\right) \\ \Rightarrow \frac{\sqrt{3}}{2} \end{gathered}$$

Hence, option B is correct.