Q73 / 180 · COMEDK 2025 (10 May Shift 3) · Mathematics
The value of $\tan \left\{\cos ^{-1}\left(\dfrac{\sqrt{2}}{2}\right)-\dfrac{\pi}{2}\right\}$ is
- A$1$
- B$\dfrac{1}{\sqrt{2}}$
- C$-\dfrac{1}{\sqrt{2}}$
- Correct answerD$-1$
Solution
The given expression is $\tan \left\{\cos ^{-1}\left(\dfrac{\sqrt{2}}{2}\right)-\dfrac{\pi}{2}\right\}$.
First, simplify the term inside the cosine inverse function: $\dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$.
We know that $\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$.
Substituting this into the expression, we get $\tan \left(\dfrac{\pi}{4} - \dfrac{\pi}{2}\right)$.
Calculating the value inside the tangent function: $\dfrac{\pi}{4} - \dfrac{\pi}{2} = -\dfrac{\pi}{4}$.
Therefore, the expression becomes $\tan \left(-\dfrac{\pi}{4}\right)$.
Since $\tan(-\theta) = -\tan(\theta)$, we have $\tan \left(-\dfrac{\pi}{4}\right) = -\tan \left(\dfrac{\pi}{4}\right) = -1$.
Answer: $-1$
First, simplify the term inside the cosine inverse function: $\dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$.
We know that $\cos^{-1}\left(\dfrac{1}{\sqrt{2}}\right) = \dfrac{\pi}{4}$.
Substituting this into the expression, we get $\tan \left(\dfrac{\pi}{4} - \dfrac{\pi}{2}\right)$.
Calculating the value inside the tangent function: $\dfrac{\pi}{4} - \dfrac{\pi}{2} = -\dfrac{\pi}{4}$.
Therefore, the expression becomes $\tan \left(-\dfrac{\pi}{4}\right)$.
Since $\tan(-\theta) = -\tan(\theta)$, we have $\tan \left(-\dfrac{\pi}{4}\right) = -\tan \left(\dfrac{\pi}{4}\right) = -1$.
Answer: $-1$
Hence, the correct answer is option: 'D'.