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Q6:
Write the following functions in the simplest form: $\tan^{-1} \frac{x}{\sqrt{a^2 - x^2}}$, $|x| < a$
Solution :
Step 1: Identifying the Trigonometric Substitution
To simplify the expression $\tan^{-1} \left( \frac{x}{\sqrt{a^2 - x^2}} \right)$, we observe the form $\sqrt{a^2 - x^2}$, which suggests a substitution based on the Pythagorean identity $\cos^2 \theta + \sin^2 \theta = 1$, specifically $1 - \sin^2 \theta = \cos^2 \theta$.
Let $x = a \sin \theta$. Given the constraint $|x| < a$, we have $\left| \frac{x}{a} \right| < 1$, which implies $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$.
[By the substitution method for inverse trigonometric integrals and simplifications]Step 2: Substitution and Algebraic Simplification
Substitute $x = a \sin \theta$ into the given expression:
$\tan^{-1} \left( \frac{a \sin \theta}{\sqrt{a^2 - (a \sin \theta)^2}} \right)$Factor out $a^2$ within the square root:
$= \tan^{-1} \left( \frac{a \sin \theta}{\sqrt{a^2(1 - \sin^2 \theta)}} \right)$Apply the trigonometric identity $1 - \sin^2 \theta = \cos^2 \theta$:
$= \tan^{-1} \left( \frac{a \sin \theta}{\sqrt{a^2 \cos^2 \theta}} \right)$ [Per the Identity: $\sin^2 \theta + \cos^2 \theta = 1$]Step 3: Evaluating the Square Root and Tangent Conversion
Since $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $\cos \theta > 0$. Therefore, $\sqrt{a^2 \cos^2 \theta} = a \cos \theta$ (assuming $a > 0$):
$= \tan^{-1} \left( \frac{a \sin \theta}{a \cos \theta} \right)$ $= \tan^{-1} (\tan \theta)$ [By the quotient identity: $\tan \theta = \frac{\sin \theta}{\cos \theta}$]Step 4: Final Domain Analysis
Since $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$, the inverse function property $\tan^{-1}(\tan \theta) = \theta$ holds true.
From the initial substitution $x = a \sin \theta$, we solve for $\theta$:
$\frac{x}{a} = \sin \theta \implies \theta = \sin^{-1} \left( \frac{x}{a} \right)$Final Solution: \sin^{-1} \left( \frac{x}{a} \right)
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: $\tan^{-1} (\frac{\cos x - \sin x}{\cos x + \sin x})$, $\frac{-\pi}{4} < x < \frac{3\pi}{4}$
- 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: $\tan^{-1} [2\cos(2\sin^{-1}(\frac{1}{2}))]$
- 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 $xy < 1$
- 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
Chapters in CBSE - Class 12 Mathematics
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