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
\[\begin{align}
  & \text{A}.1 \\
 & \text{B}.2 \\
 & \text{C}\text{.}\sqrt{2} \\
 & \text{D}\text{.}\sqrt{3} \\
\end{align}\]

Answer 1

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.

Complete step-by-step answer:
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).