Inverse Trigonometric Functions
NCERT Class 12
CBSE
Grade 12
Question:
Express $\tan^{-1}\left(\dfrac{\cos x}{1 - \sin x}\right)$, $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$ in the simplest form.
Step-by-Step Solution
$\dfrac{\cos x}{1 - \sin x} = \dfrac{\cos^2(x/2) - \sin^2(x/2)}{(\cos(x/2) - \sin(x/2))^2} = \dfrac{\cos(x/2) + \sin(x/2)}{\cos(x/2) - \sin(x/2)} = \tan(\pi/4 + x/2)$. [1.0 Mark]
$\tan^{-1}(\tan(\pi/4 + x/2)) = \dfrac{\pi}{4} + \dfrac{x}{2}$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Transforming argument to $\tan(\pi/4 + x/2)$: 1.0 Mark
Evaluating simplest form $= \pi/4 + x/2$: 1.0 Mark
Correct Answer:
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