Question

The value of $\tan \left(67{\frac{1}{2}}^{°}\right)+\cot \left(67{\frac{1}{2}}^{°}\right)$ is

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

Answer 1

The correct option is C

$2\sqrt{2}$

Explanation for the correct option

The given trigonometric expression: $\tan \left(67{\frac{1}{2}}^{°}\right)+\cot \left(67{\frac{1}{2}}^{°}\right)$.

It is known that, $\mathbf{tan}\left(180°-\theta \right)\mathbf{=}\mathbf{-}\mathbf{tan}\left(\theta \right)$

Thus, $\tan \left(135°\right)=\tan \left(180°-45°\right)$

$$\begin{gathered} \Rightarrow \tan \left(135°\right)=-\tan \left(45°\right) \\ \Rightarrow \tan \left(135°\right)=-1 \end{gathered}$$.

Now, $\mathbf{tan}\mathbf{\left(}\mathbf{2}\mathbf{\theta }\mathbf{\right)}\mathbf{=}\frac{\mathbf{2}\mathbf{tan}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}}{\mathbf{1}\mathbf{-}{\mathbf{tan}}^{\mathbf{2}}\left(\theta \right)}$.

Thus, $\tan \left(135°\right)=\frac{2\tan \left({{\displaystyle \frac{135}{2}}}^{°}\right)}{1-{\tan }^2\left({\frac{135}{2}}^{°}\right)}$.

$$\begin{gathered} \Rightarrow -1=\frac{2\tan \left({{\displaystyle \frac{135}{2}}}^{°}\right)}{1-{\tan }^2\left({\frac{135}{2}}^{°}\right)} \\ \Rightarrow -1+{\tan }^2\left({\frac{135}{2}}^{°}\right)=2\tan \left({\frac{135}{2}}^{°}\right) \\ \Rightarrow {\tan }^2\left({\frac{135}{2}}^{°}\right)-2\tan \left({\frac{135}{2}}^{°}\right)-1=0 \\ \Rightarrow {\tan }^2\left(67{\frac{1}{2}}^{°}\right)-2\tan \left(67{\frac{1}{2}}^{°}\right)-1=0 \end{gathered}$$

Apply the quadratic formula to solve the equation, $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$.

$$\begin{gathered} \tan \left(67{\frac{1}{2}}^{°}\right)=\frac{-\left(-2\right)\pm \sqrt{{\left(-2\right)}^2-4\left(1\right)\left(-1\right)}}{2\left(1\right)} \\ \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)=\frac{2\pm \sqrt{4+4}}{2} \\ \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)=\frac{2\pm \sqrt{8}}{2} \\ \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)=\frac{2\pm 2\sqrt{2}}{2} \\ \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)=1\pm \sqrt{2} \end{gathered}$$

As $67{\frac{1}{2}}^{°}<90°$, thus $\tan \left(67{\frac{1}{2}}^{°}\right)>0$.

So, $\tan \left(67{\frac{1}{2}}^{°}\right)=1+\sqrt{2}$

Therefore, $\tan \left(67{\frac{1}{2}}^{°}\right)+\cot \left(67{\frac{1}{2}}^{°}\right)=\tan \left(67{\frac{1}{2}}^{°}\right)+\frac{1}{\tan \left(67{\frac{1}{2}}^{°}\right)}\left[\because \cot \left(x\right)=\frac{1}{\tan \left(x\right)}\right]$

$$\begin{gathered} \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)+\cot \left(67{\frac{1}{2}}^{°}\right)=1+\sqrt{2}+\frac{1}{1+\sqrt{2}} \\ \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)+\cot \left(67{\frac{1}{2}}^{°}\right)=1+\sqrt{2}+\frac{\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)} \\ \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)+\cot \left(67{\frac{1}{2}}^{°}\right)=1+\sqrt{2}+\frac{\left(1-\sqrt{2}\right)}{{\left(1\right)}^2-{\left(\sqrt{2}\right)}^2} \\ \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)+\cot \left(67{\frac{1}{2}}^{°}\right)=1+\sqrt{2}+\frac{\left(1-\sqrt{2}\right)}{1-2} \\ \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)+\cot \left(67{\frac{1}{2}}^{°}\right)=1+\sqrt{2}-1+\sqrt{2} \\ \Rightarrow \tan \left(67{\frac{1}{2}}^{°}\right)+\cot \left(67{\frac{1}{2}}^{°}\right)=2\sqrt{2} \end{gathered}$$

Hence, option C is correct .