Question

The value of $\tan {{20}^{\circ }}+2\tan {{50}^{\circ }}-\tan {{70}^{\circ }}$
$\begin{aligned} & A)1 \\\ & B)0 \\\ & C)\tan {{50}^{\circ }} \\\ \end{aligned}$
D)None of these

Answer 1

The above question can be solved by basic definition of trigonometry and using the various identities involved it, the following identities can be used to solve this problem:
$\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$
$\tan \theta =\cot \left( 90-\theta \right)$
So, we start by expressing $\tan {{70}^{\circ }}=\tan \left( {{20}^{\circ }}+{{50}^{\circ }} \right)$ and then applying the $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$.

Complete step by step solution:
Now write the expression and use simple trigonometric results to it:
$\tan {{20}^{\circ }}+2\tan {{50}^{\circ }}-\tan {{70}^{\circ }}$
Now we can write $\tan {{70}^{\circ }}=\tan \left( {{20}^{\circ }}+{{50}^{\circ }} \right)$
The above expression can be written by using the value listed above, we get:
\Rightarrow $$$\tan {{20}^{\circ }}+2\tan {{50}^{\circ }}-\tan \left( {{20}^{\circ }}+{{50}^{\circ }} \right)$ Now using the property of trigonometry for the expression $\tan {{70}^{\circ }}=\tan \left( {{20}^{\circ }}+{{50}^{\circ }} \right)$ we use the property $\tan \left( A+B \right)=\dfrac{\tan A+\tan B}{1-\tan A\tan B}$ we get: \Rightarrow $$$\tan \left( {{20}^{\circ }}+{{50}^{\circ }} \right)=\dfrac{\tan {{20}^{\circ }}+\tan {{50}^{\circ }}}{1-\tan {{20}^{\circ }}\tan {{50}^{\circ }}}$Multiplying the denominator term to left-hand side in the above expression and substituting$\tan {{70}^{\circ }}=\tan \left( {{20}^{\circ }}+{{50}^{\circ }} \right)we get: $$\Rightarrow $$$$\tan {{70}^{\circ }}\left( 1-\tan {{20}^{\circ }}\tan {{50}^{\circ }} \right)=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}$$ Multiplying each term in the bracket by\tan {{70}^{\circ }}we get: $$\Rightarrow \tan {{70}^{\circ }}-\tan {{20}^{\circ }}\tan {{70}^{\circ }}\tan {{50}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}$$ Since we know that\tan \theta =\cot \left( 90-\theta \right)$therefore we can write:$\begin{aligned}
& \tan {{70}^{\circ }}=\tan \left( {{90}^{\circ }}-{{70}^{\circ }} \right) \\
& \Rightarrow \tan {{70}^{\circ }}=\cot {{20}^{\circ }} \\
\end{aligned} Using this the above expression can be written as: $$\Rightarrow \tan {{70}^{\circ }}-\tan {{20}^{\circ }}\cot {{20}^{\circ }}\tan {{50}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }}$$ Now we can use the property that\tan \theta =\dfrac{1}{\cot \theta }$we can write the above expression as follows. Substitute the value of$\tan \theta $in terms of$\cot \theta $, we get:
$\Rightarrow \tan {{70}^{\circ }}-\tan {{20}^{\circ }}\times \dfrac{1}{\tan {{20}^{\circ }}}\tan {{50}^{\circ }}=\tan {{20}^{\circ }}+\tan {{70}^{\circ }}$
The expression becomes:

& \Rightarrow \tan {{70}^{\circ }}-\tan {{50}^{\circ }}=\tan {{20}^{\circ }}+\tan {{50}^{\circ }} \\\ & \Rightarrow \tan {{70}^{\circ }}=\tan {{20}^{\circ }}+2\tan {{50}^{\circ }} \\\ \end{aligned}$$ Now to find the value of $\tan {{20}^{\circ }}+2\tan {{50}^{\circ }}-\tan {{70}^{\circ }}$ We know that $$\tan {{70}^{\circ }}=\tan {{20}^{\circ }}+2\tan {{50}^{\circ }}$$ therefore substitute its value in equation, we get: $$\begin{aligned} & \tan {{20}^{\circ }}+2\tan {{50}^{\circ }}-\tan {{20}^{\circ }}+2\tan {{50}^{\circ }} \\\ & \therefore 0 \\\ \end{aligned}$$ **So, the correct answer is “Option (B)”.** **Note:** In order to solve this question we can make use of the identities or we can also put the values of various terms involved in expression from the calculator we find that the value in either of the cases is 0. It should be kept in mind that we don’t use an identity that could make the expression more complex instead a simple identity has to be used in order to have a flexible way to solve the problem.