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Q9:
Find the values of each of the following: $\tan \frac{1}{2}[\sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2}]$, $|x| < 1$, $y > 0$ and $xy < 1$

Solution :

Step 1: Application of Substitution Identities

To simplify the expression $\tan \frac{1}{2}[\sin^{-1} \frac{2x}{1+x^2} + \cos^{-1} \frac{1-y^2}{1+y^2}]$, we utilize the standard trigonometric substitution methods [Per the substitution $x = \tan \theta$].

Recall the standard inverse trigonometric identities:

  • $\sin^{-1} \left( \frac{2x}{1+x^2} \right) = 2 \tan^{-1} x$, provided $|x| \le 1$
  • $\cos^{-1} \left( \frac{1-y^2}{1+y^2} \right) = 2 \tan^{-1} y$, provided $y \ge 0$

Step 2: Substitution into the Original Expression

Substitute the derived identities into the expression:

$\text{Expression} = \tan \frac{1}{2} \left[ 2 \tan^{-1} x + 2 \tan^{-1} y \right]$

Distribute the $\frac{1}{2}$ across the brackets:

$\text{Expression} = \tan \left( \tan^{-1} x + \tan^{-1} y \right)$

Step 3: Application of Inverse Tangent Addition Formula

Utilize the identity for the sum of inverse tangents [Per the Addition Formula for Inverse Trigonometric Functions]:

$\tan^{-1} x + \tan^{-1} y = \tan^{-1} \left( \frac{x+y}{1-xy} \right)$

Since the problem states $xy < 1$, the condition for the standard addition formula is satisfied. Substituting this into our expression:

$\text{Expression} = \tan \left( \tan^{-1} \left( \frac{x+y}{1-xy} \right) \right)$

Step 4: Simplification via Property of Inverse Functions

Given the property $\tan(\tan^{-1} \theta) = \theta$ for $\theta \in \mathbb{R}$, we simplify the expression directly:

$\text{Expression} = \frac{x+y}{1-xy}$

Visual Representation of the Domain Mapping

Input variable space (x, y) Output: f(x,y) y = (x+y)/(1-xy)

Final Solution: \frac{x+y}{1-xy}


More Questions from Class 12 Mathematics Inverse Trigonometric Functions EXERCISE 2.2


CBSE Solutions for Class 12 Mathematics Inverse Trigonometric Functions


Chapters in CBSE - Class 12 Mathematics


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