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Question: If the value of \[\dfrac{\text{tan2}{{\text{6}}^{\circ }}+\text{tan1}{{\text{9}}^{\circ }}}{x\left( ...

If the value of tan26∘+tan19∘x(1−tan26∘+tan19∘)=cos60∘\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

& \text{A}.1 \\\ & \text{B}.2 \\\ & \text{C}\text{.}\sqrt{2} \\\ & \text{D}\text{.}\sqrt{3} \\\ \end{aligned}$$
Explanation

Solution

To solve this question, we will use the trigonometric identity which is given as below:
tanA+tanB1-tanAtanB=tan(A+B)\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∘{{26}^{\circ }} and B as 19∘{{19}^{\circ }} after applying this identity, we will simplify by using the fact that, tan45∘=1 and cos60∘=12\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 tan26∘+tan19∘x(1−tan26∘+tan19∘)=cos60∘ . . . . . . . . . . . . . . (i)\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:
tanA+tanB1-tanAtanB=tan(A+B)\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 A=26∘\text{A=}{{26}^{\circ }} and the value of B=19∘\text{B=}{{19}^{\circ }}
Applying the identity stated above and using A=26∘ and B=19∘\text{A=}{{26}^{\circ }}\text{ and B=}{{19}^{\circ }} we get:
tan(26∘+19∘)=tan26∘+tan19∘1-tan26∘+tan19∘ . . . . . . . . . . . . (ii)\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:
tan26∘+tan19∘1-tan26∘+tan19∘=xcos60∘ . . . . . . . . . . . . . . (iii)\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:

& \text{xcos}{{60}^{\circ }}\text{=tan}\left( {{26}^{\circ }}+{{19}^{\circ }} \right) \\\ & \Rightarrow \text{tan}\left( {{45}^{\circ }} \right)\text{=xcos}{{60}^{\circ }} \\\ \end{aligned}$$ We know that, the value of $$\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).