Inverse Trigonometric Functions
NCERT Class 12
CBSE
Grade 12
Question:
Find the value of $\tan\left(\dfrac{1}{2} \sin^{-1} \dfrac{2x}{1 + x^2} + \dfrac{1}{2} \cos^{-1} \dfrac{1 - y^2}{1 + y^2}\right)$, $|x| < 1, y > 0$ and $xy < 1$.
Step-by-Step Solution
Use identities: $\frac{1}{2}(2\tan^{-1}x) + \frac{1}{2}(2\tan^{-1}y) = \tan^{-1}x + \tan^{-1}y$. [1.5 Marks]
$\tan(\tan^{-1}((x+y)/(1-xy))) = \dfrac{x + y}{1 - xy}$. [1.5 Marks]
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🎯 Official CBSE Marking Scheme:
Applying $2\tan^{-1}$ identities: 1.5 Marks
Evaluating value $= (x+y)/(1-xy)$: 1.5 Marks
Correct Answer:
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