Question

The value of $\tan \left(20°\right)+2\tan \left(50°\right)-\tan \left(70°\right)$ is

• A. $1$
• B. $0$
• C. $\tan \left(50°\right)$
• D. None of these.

Answer 1

The correct option is B

$0$

Explanation of the correct option

The given trigonometric expression: $\tan \left(20°\right)+2\tan \left(50°\right)-\tan \left(70°\right)$.

$$\begin{gathered} \tan \left(50°\right)=\tan \left(70°-20°\right) \\ \Rightarrow \tan \left(50°\right)=\frac{\tan \left(70°\right)-\tan \left(20°\right)}{1+\tan \left(70°\right)\tan \left(20°\right)}\left[\because \tan \left(A-B\right)=\frac{\tan \left(A\right)-\tan \left(B\right)}{1+\tan \left(A\right)\tan \left(B\right)}\right] \\ \Rightarrow \tan \left(50°\right)=\frac{\tan \left(70°\right)-\tan \left(20°\right)}{1+\tan \left(90°-20°\right)\tan \left(20°\right)} \\ \Rightarrow \tan \left(50°\right)=\frac{\tan \left(70°\right)-\tan \left(20°\right)}{1+\cot \left(20°\right)\tan \left(20°\right)}\left[\because \tan \left(90°-x\right)=\cot \left(x\right)\right] \\ \Rightarrow \tan \left(50°\right)=\frac{\tan \left(70°\right)-\tan \left(20°\right)}{1+\left[{\displaystyle \frac{1}{\tan \left(20°\right)}}\times \tan \left(20°\right)\right]}\left[\because \cot \left(x\right)=\frac{1}{\tan \left(x\right)}\right] \\ \Rightarrow \tan \left(50°\right)=\frac{\tan \left(70°\right)-\tan \left(20°\right)}{1+1} \\ \Rightarrow \tan \left(50°\right)=\frac{\tan \left(70°\right)-\tan \left(20°\right)}{2} \\ \Rightarrow 2\tan \left(50°\right)=\tan \left(70°\right)-\tan \left(20°\right) \\ \Rightarrow \tan \left(20°\right)+2\tan \left(50°\right)-\tan \left(70°\right)=0 \end{gathered}$$

Hence, option B is correct .