Question

If $\tan \left(\mathrm{\alpha }\right)+\cot \left(\mathrm{\alpha }\right)=2$ then find the value of ${\cot }^{20}\left(\mathrm{\alpha }\right)+{\tan }^{20}\left(\mathrm{\alpha }\right)$ is

• A. $0$
• B. $2$
• C. $20$
• D. $220$

Answer 1

The correct option is B

$2$

Step 1 : Simplifying using basic trigonometric identities

Use $\cot \left(\mathrm{\alpha }\right)=\frac{1}{\tan \left(\mathrm{\alpha }\right)}$

From given equation

$\begin{array}{rcl}\tan \left(\alpha \right)+\frac{1}{\tan \left(\alpha \right)}& =& 2\\ & \Rightarrow & {\tan }^2\left(\alpha \right)-2\tan \left(\alpha \right)+1=0\end{array}$

Step 2 : Find roots of Quadratic equation obtained

$$\begin{gathered} {\left(\tan \left(\mathrm{\alpha }\right)-1\right)}^2=0\left[\because {\left(\mathrm{a}-\mathrm{b}\right)}^2={\mathrm{a}}^2+{\mathrm{b}}^2-2\mathrm{ab}\right] \\ \Rightarrow \tan \left(\mathrm{\alpha }\right)=1 \\ \Rightarrow \cot \left(\mathrm{\alpha }\right)=1 \end{gathered}$$

Step 3 : Find value of given expression

$$\begin{gathered} {\cot }^{20}\left(\mathrm{\alpha }\right)+{\tan }^{20}\left(\mathrm{\alpha }\right)=1^{20}+1^{20} \\ \Rightarrow {\cot }^{20}\left(\mathrm{\alpha }\right)+{\tan }^{20}\left(\mathrm{\alpha }\right)=2 \end{gathered}$$

Hence option B is the correct answer.