Inverse Trigonometric Functions
NCERT Class 12
CBSE
Grade 12
Question:
Simplest form of $\tan^{-1}\left(\dfrac{\cos x - \sin x}{\cos x + \sin x}\right)$, $-\dfrac{\pi}{4} < x < \dfrac{3\pi}{4}$ is:
(a) $\dfrac{\pi}{4} - x$
(b) $\dfrac{\pi}{4} + x$
(c) $x - \dfrac{\pi}{4}$
(d) $\pi - x$
Step-by-Step Solution
$\tan^{-1}\left(\dfrac{1-\tan x}{1+\tan x}\right) = \tan^{-1}(\tan(\pi/4 - x)) = \pi/4 - x$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Evaluating simplest form $= \pi/4 - x$: 1.0 Mark
Correct Answer: $\dfrac{\pi}{4} - x$
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