Question

Evaluate $\tan A+\cot (180°+A)+\cot \left(90°+A\right)+\cot \left(360°-A\right)$

• A. $0$
• B. $2\tan A$
• C. $2\cot A$
• D. $2\left(\tan A-\cot A\right)$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Simplifying the given expression

Given, $\tan A+\cot (180°+A)+\cot \left(90°+A\right)+\cot \left(360°-A\right)$

We know that:

$$\begin{gathered} \cot \left(180+x\right)=\cot \left(x\right) \\ \cot \left(90+x\right)=-\tan \left(x\right) \\ \cot \left(360-x\right)=-\cot \left(x\right) \end{gathered}$$

On solving we get,

$$\begin{gathered} \tan A+\cot (180°+A)+\cot \left(90°+A\right)+\cot \left(360°-A\right)=\tan A+\mathrm{co}tA-\tan A-\cot A \\ =0 \end{gathered}$$

Hence, the correct answer is Option (A).