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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
Final Solution: \frac{x+y}{1-xy}
More Questions from Class 12 Mathematics Inverse Trigonometric Functions EXERCISE 2.2
- Q1: Prove the following: $3\sin^{-1} x = \sin^{-1} (3x - 4x^3)$, $x \in [\frac{-1}{2}, \frac{1}{2}]$
- Q2: Prove the following: $3\cos^{-1} x = \cos^{-1} (4x^3 - 3x)$, $x \in [\frac{1}{2}, 1]$
- Q3: Write the following functions in the simplest form: $\tan^{-1} \frac{\sqrt{1+x^2}-1}{x}$, $x \neq 0$
- Q4: Write the following functions in the simplest form: $\tan^{-1} (\sqrt{\frac{1-\cos x}{1+\cos x}})$, $0 < x < \pi$
- Q5: Write the following functions in the simplest form: $\tan^{-1} (\frac{\cos x - \sin x}{\cos x + \sin x})$, $\frac{-\pi}{4} < x < \frac{3\pi}{4}$
- Q6: Write the following functions in the simplest form: $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}$, $|x| < a$
- Q7: Write the following functions in the simplest form: $\tan^{-1} (\frac{3a^2x - x^3}{a^3 - 3ax^2})$, $a > 0$; $\frac{-a}{\sqrt{3}} < x < \frac{a}{\sqrt{3}}$
- Q8: Find the values of each of the following: $\tan^{-1} [2\cos(2\sin^{-1}(\frac{1}{2}))]$
- Q10: Find the values of each of the expressions: $\sin^{-1}(\sin(\frac{2\pi}{3}))$
- Q11: Find the values of each of the expressions: $\tan^{-1}(\tan(\frac{3\pi}{4}))$
- Q12: Find the values of each of the expressions: $\tan(\sin^{-1}(\frac{3}{5}) + \cot^{-1}(\frac{3}{2}))$
- Q13: $\cos^{-1}(\cos(\frac{7\pi}{6}))$ is equal to
- Q14: $\sin(\frac{\pi}{3} - \sin^{-1}(\frac{-1}{2}))$ is equal to
- Q15: $\tan^{-1}(\sqrt{3}) - \cot^{-1}(-\sqrt{3})$ is equal to
CBSE Solutions for Class 12 Mathematics Inverse Trigonometric Functions
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