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Trigonometric Functions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsTrigonometric Functions1 Mark
MCQ
The smallest value of $x$ satisfying the equation $\sqrt{3}(\cot\text{x}+\tan\text{x})=4$ is:
  • A
    $\frac{2\pi}{3}$
  • B
    $\frac{\pi}{3}$
  • ✓
    $\frac{\pi}{6}$
  • D
    $\frac{\pi}{12}$
✓

Answer

Correct option: C.
$\frac{\pi}{6}$
Given:
$\sqrt{3}(\cot\text{x}+\tan\text{x})=4$
$\Rightarrow\sqrt{3}\Big(\frac{\cos\text{x}}{\sin\text{x}}+\frac{\sin\text{x}}{\cos\text{x}}\Big)=4$
$\Rightarrow\sqrt{3}(\cos^2\text{x}+\sin^2\text{x})=4\sin\text{x}\cos\text{x}$
$\Rightarrow\sqrt{3}=2\sin2\text{x}$ $\big[\sin2\text{x}=2\sin\text{x}\cos\text{x}\big]$
$\Rightarrow\sin2\text{x}=\frac{\sqrt{3}}{2}$
$\Rightarrow\sin2\text{x}=\sin\frac{\pi}{3}$
$\Rightarrow2\text{x}=\text{n}\pi+(-1)^{\text{n}}\frac{\pi}{3},\text{n}\in\text{Z}$
$\Rightarrow\text{x}=\frac{\text{n}\pi}{2}+(-1)^{\text{n}}\frac{\pi}{6},\text{n}\in\text{Z}$
To obtain the smallest value of $x,$ we will put $n = 0n = 0$ in the above equation.
Thus, we have:
$\text{x}=\frac{\pi}{6}$
Hence, the smallest value of $x$ is $\frac{\pi}{6}.$

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