Question

# If the value of $\dfrac{\text{tan2}{{\text{6}}^{\circ }}+\text{tan1}{{\text{9}}^{\circ }}}{x\left( 1-\text{tan}{{26}^{\circ }}+\text{tan}{{19}^{\circ }} \right)}=\text{cos}{{60}^{\circ }}$ then the value of x is\begin{align} & \text{A}.1 \\ & \text{B}.2 \\ & \text{C}\text{.}\sqrt{2} \\ & \text{D}\text{.}\sqrt{3} \\ \end{align}

Hint:
To solve this question, we will use the trigonometric identity which is given as below:
$\dfrac{\text{tanA+tanB}}{\text{1-tanAtanB}}\text{=tan}\left( \text{A+B} \right)$
Where A and B are angles. In our case, we take A as ${{26}^{\circ }}$ and B as ${{19}^{\circ }}$ after applying this identity, we will simplify by using the fact that, $\text{tan}{{45}^{\circ }}=1\text{ and cos}{{60}^{\circ }}=\dfrac{1}{2}$ to get the value of x.

Given that $\dfrac{\text{tan2}{{\text{6}}^{\circ }}+\text{tan1}{{\text{9}}^{\circ }}}{x\left( 1-\text{tan}{{26}^{\circ }}+\text{tan}{{19}^{\circ }} \right)}=\text{cos}{{60}^{\circ }}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}$
To solve this question, we will use the trigonometric identity which is given as below:
$\dfrac{\text{tanA+tanB}}{\text{1-tanAtanB}}\text{=tan}\left( \text{A+B} \right)$
Where A and B are angles.
To solve this question, let us assume $\text{A=}{{26}^{\circ }}$ and the value of $\text{B=}{{19}^{\circ }}$
Applying the identity stated above and using $\text{A=}{{26}^{\circ }}\text{ and B=}{{19}^{\circ }}$ we get:
$\text{tan}\left( {{26}^{\circ }}+{{19}^{\circ }} \right)\text{=}\dfrac{\text{tan}{{26}^{\circ }}\text{+tan}{{19}^{\circ }}}{\text{1-tan}{{26}^{\circ }}\text{+tan}{{19}^{\circ }}}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}$
In equation (i) multiplying both sides by x, we get:
$\dfrac{\text{tan}{{26}^{\circ }}\text{+tan}{{19}^{\circ }}}{\text{1-tan}{{26}^{\circ }}\text{+tan}{{19}^{\circ }}}\text{=xcos}{{60}^{\circ }}\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (iii)}$
Comparing from equation (ii) and (iii) we get:
\begin{align} & \text{xcos}{{60}^{\circ }}\text{=tan}\left( {{26}^{\circ }}+{{19}^{\circ }} \right) \\ & \Rightarrow \text{tan}\left( {{45}^{\circ }} \right)\text{=xcos}{{60}^{\circ }} \\ \end{align}
We know that, the value of $\text{tan}{{45}^{\circ }}=1\text{ and cos}{{60}^{\circ }}=\dfrac{1}{2}$
Substituting this above, we get:
$\text{x}\dfrac{1}{2}=1$
Multiply both sides with 2, we get:
$\Rightarrow \text{x=2}$
Hence the value of x = 2

So, the correct answer is “Option B”.

Note: The key point to note in this question is that, the trigonometric identity $\dfrac{\text{tanA+tanB}}{\text{1-tanAtanB}}\text{=tan}\left( \text{A+B} \right)$ is applicable when we have to calculate tan (A + B).