Inverse Trigonometric Functions
NCERT Class 12
CBSE
Grade 12
Question:
The value of $\tan^{-1}(\sqrt{3}) - \sec^{-1}(-2)$ is:
(a) $-\dfrac{\pi}{3}$
(b) $\dfrac{\pi}{3}$
(c) $\pi$
(d) $\dfrac{2\pi}{3}$
Step-by-Step Solution
$\tan^{-1}(\sqrt{3}) = \pi/3, \sec^{-1}(-2) = 2\pi/3 \Rightarrow \pi/3 - 2\pi/3 = -\pi/3$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating expression $= -\pi/3$: 1.0 Mark
Correct Answer: $-\dfrac{\pi}{3}$
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