Q105 / 180 · COMEDK 2024 (Morning Shift) · Mathematics
$\text { The value of } \sin ^{-1}\left[\cot \left(\dfrac{1}{2} \tan ^{-1} \dfrac{1}{\sqrt{3}}+\cos ^{-1} \dfrac{\sqrt{12}}{4}+\sin ^{-1} \dfrac{1}{\sqrt{2}}\right)\right]
$ is
- A$\dfrac{\pi}{6}$
- Correct answerB0
- C$\dfrac{\pi}{4}$
- D$\dfrac{\pi}{2}$
Solution
Let the expression inside the cotangent be $\theta$.
$\theta = \dfrac{1}{2} \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) + \cos^{-1} \left(\dfrac{\sqrt{12}}{4}\right) + \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right)$
We know that $\tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$.
Also, $\dfrac{\sqrt{12}}{4} = \dfrac{2\sqrt{3}}{4} = \dfrac{\sqrt{3}}{2}$, so $\cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{6}$.
And $\sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$.
Substituting these values into $\theta$:
$\theta = \dfrac{1}{2} \left(\dfrac{\pi}{6}\right) + \dfrac{\pi}{6} + \dfrac{\pi}{4} = \dfrac{\pi}{12} + \dfrac{\pi}{6} + \dfrac{\pi}{4}$
Finding a common denominator of 12:
$\theta = \dfrac{\pi + 2\pi + 3\pi}{12} = \dfrac{6\pi}{12} = \dfrac{\pi}{2}$
The expression becomes $\sin^{-1} [\cot(\theta)] = \sin^{-1} [\cot(\pi/2)]$.
Since $\cot(\pi/2) = 0$, we have $\sin^{-1}(0) = 0$.
Answer: 0
$\theta = \dfrac{1}{2} \tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) + \cos^{-1} \left(\dfrac{\sqrt{12}}{4}\right) + \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right)$
We know that $\tan^{-1} \left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6}$.
Also, $\dfrac{\sqrt{12}}{4} = \dfrac{2\sqrt{3}}{4} = \dfrac{\sqrt{3}}{2}$, so $\cos^{-1} \left(\dfrac{\sqrt{3}}{2}\right) = \dfrac{\pi}{6}$.
And $\sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$.
Substituting these values into $\theta$:
$\theta = \dfrac{1}{2} \left(\dfrac{\pi}{6}\right) + \dfrac{\pi}{6} + \dfrac{\pi}{4} = \dfrac{\pi}{12} + \dfrac{\pi}{6} + \dfrac{\pi}{4}$
Finding a common denominator of 12:
$\theta = \dfrac{\pi + 2\pi + 3\pi}{12} = \dfrac{6\pi}{12} = \dfrac{\pi}{2}$
The expression becomes $\sin^{-1} [\cot(\theta)] = \sin^{-1} [\cot(\pi/2)]$.
Since $\cot(\pi/2) = 0$, we have $\sin^{-1}(0) = 0$.
Answer: 0
Hence, the correct answer is option: 'B'.