Question

The value of $\tan \left(7{\frac{1}{2}}^{°}\right)$ is equal to

• A. $\sqrt{6}+\sqrt{3}+\sqrt{2}-2$
• B. $\sqrt{6}-\sqrt{3}+\sqrt{2}-2$
• C. $\sqrt{6}-\sqrt{3}+\sqrt{2}+2$
• D. $\sqrt{6}-\sqrt{3}-\sqrt{2}-2$

Answer 1

The correct option is B

$\sqrt{6}-\sqrt{3}+\sqrt{2}-2$

Explanation for the correct option

The given trigonometric expression: $\tan \left(7{\frac{1}{2}}^{°}\right)$.

It is known that $\mathbf{sin}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{-}\mathbf{cos}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\mathbf{B}\mathbf{\right)}$.

Thus, $\sin \left(15°\right)=\sin \left(45°-30°\right)$

$$\begin{gathered} \Rightarrow \sin \left(15°\right)=\sin \left(45°\right)\cos \left(30°\right)-\cos \left(45°\right)\sin \left(30°\right) \\ \Rightarrow \sin \left(15°\right)=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\times \frac{1}{2} \\ \Rightarrow \sin \left(15°\right)=\frac{\sqrt{3}}{2\sqrt{2}}-\frac{1}{2\sqrt{2}} \\ \Rightarrow \sin \left(15°\right)=\frac{\sqrt{3}-1}{2\sqrt{2}} \\ \Rightarrow \sin \left(15°\right)=\frac{\left(2\sqrt{2}\right)\left(\sqrt{3}-1\right)}{{\left(2\sqrt{2}\right)}^2} \\ \Rightarrow \sin \left(15°\right)=\frac{2\left(\sqrt{6}-\sqrt{2}\right)}{8} \\ \Rightarrow \sin \left(15°\right)=\frac{\left(\sqrt{6}-\sqrt{2}\right)}{4} \end{gathered}$$

It is known that $\mathbf{cos}\mathbf{(}\mathbf{A}\mathbf{-}\mathbf{B}\mathbf{)}\mathbf{=}\mathbf{cos}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{cos}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{+}\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{sin}\mathbf{\left(}\mathbf{B}\mathbf{\right)}$.

Thus, $\cos \left(15°\right)=\cos \left(45°-30°\right)$

$$\begin{gathered} \Rightarrow \cos \left(15°\right)=\cos \left(45°\right)\cos \left(30°\right)+\sin \left(45°\right)\sin \left(30°\right) \\ \Rightarrow \cos \left(15°\right)=\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2} \\ \Rightarrow \cos \left(15°\right)=\frac{\sqrt{3}}{2\sqrt{2}}+\frac{1}{2\sqrt{2}} \\ \Rightarrow \cos \left(15°\right)=\frac{\sqrt{3}+1}{2\sqrt{2}} \\ \Rightarrow \cos \left(15°\right)=\frac{\left(2\sqrt{2}\right)\left(\sqrt{3}+1\right)}{{\left(2\sqrt{2}\right)}^2} \\ \Rightarrow \cos \left(15°\right)=\frac{2\left(\sqrt{6}+\sqrt{2}\right)}{8} \\ \Rightarrow \cos \left(15°\right)=\frac{\left(\sqrt{6}+\sqrt{2}\right)}{4} \end{gathered}$$

Now, $\tan \left(A\right)=\frac{\sin \left(A\right)}{\cos \left(A\right)}$.

$$\begin{gathered} \Rightarrow \tan \left(A\right)=\frac{2\sin \left(A\right)\cos \left(A\right)}{2{\cos }^2\left(A\right)} \\ \Rightarrow \tan \left(A\right)=\frac{\sin \left(2A\right)}{1+\cos \left(2A\right)} \end{gathered}$$

Thus, $\tan \left(7{\frac{1}{2}}^{°}\right)=\frac{\sin \left(15°\right)}{1+\cos \left(15°\right)}$.

$$\begin{gathered} \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\frac{{\displaystyle \frac{\sqrt{6}-\sqrt{2}}{4}}}{1+\frac{\sqrt{6}+\sqrt{2}}{4}} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\frac{{\displaystyle \sqrt{6}-\sqrt{2}}}{4+\sqrt{6}+\sqrt{2}} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\frac{{\displaystyle \left(\sqrt{6}-\sqrt{2}\right)\left[4-\left(\sqrt{6}+\sqrt{2}\right)\right]}}{\left[4+\left(\sqrt{6}+\sqrt{2}\right)\right]\left[4-\left(\sqrt{6}+\sqrt{2}\right)\right]} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\frac{4\left(\sqrt{6}-\sqrt{2}\right)-\left(\sqrt{6}-\sqrt{2}\right)\left(\sqrt{6}+\sqrt{2}\right)}{16-{\left(\sqrt{6}+\sqrt{2}\right)}^2} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\frac{4\sqrt{6}-4\sqrt{2}-6+2}{16-6-2-2\sqrt{12}} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\frac{4\sqrt{6}-4\sqrt{2}-4}{8-4\sqrt{3}} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\frac{\sqrt{6}-\sqrt{2}-1}{2-\sqrt{3}} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\frac{\left(\sqrt{6}-\sqrt{2}-1\right)\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\frac{2\sqrt{6}-2\sqrt{2}-2+\sqrt{18}-\sqrt{6}-\sqrt{3}}{4-3} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\sqrt{6}-2\sqrt{2}-2+3\sqrt{2}-\sqrt{3} \\ \Rightarrow \tan \left(7{\frac{1}{2}}^{°}\right)=\sqrt{6}-\sqrt{3}+\sqrt{2}-2 \end{gathered}$$

Hence, option B is correct.