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Q5:
Write the following functions in the simplest form: $\tan^{-1} (\frac{\cos x - \sin x}{\cos x + \sin x})$, $\frac{-\pi}{4} < x < \frac{3\pi}{4}$
Solution :
Step 1: Algebraic Manipulation of the Trigonometric Expression
Given the expression: $f(x) = \tan^{-1} \left( \frac{\cos x - \sin x}{\cos x + \sin x} \right)$.
To simplify the argument of the inverse tangent function, divide both the numerator and the denominator by $\cos x$ [Justification: This transformation is valid as $\cos x \neq 0$ within the domain $\frac{-\pi}{4} < x < \frac{3\pi}{4}$].
$$\frac{\frac{\cos x}{\cos x} - \frac{\sin x}{\cos x}}{\frac{\cos x}{\cos x} + \frac{\sin x}{\cos x}} = \frac{1 - \tan x}{1 + \tan x}$$
Step 2: Applying Trigonometric Identities
Recall the trigonometric identity for the tangent of a difference: $\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}$.
We can rewrite the expression $1 - \tan x$ by substituting $1$ with $\tan \frac{\pi}{4}$:
$$\frac{\tan \frac{\pi}{4} - \tan x}{1 + (\tan \frac{\pi}{4})(\tan x)}$$
[Per the Addition/Subtraction Formula for Tangent, this simplifies to $\tan(\frac{\pi}{4} - x)$].
Step 3: Simplification via Inverse Trigonometric Properties
Now, the function becomes: $f(x) = \tan^{-1} \left( \tan \left( \frac{\pi}{4} - x \right) \right)$.
We must check if the argument $(\frac{\pi}{4} - x)$ lies within the principal value branch of $\tan^{-1} x$, which is $(-\frac{\pi}{2}, \frac{\pi}{2})$.
- Given: $-\frac{\pi}{4} < x < \frac{3\pi}{4}$
- Multiply by $-1$: $-\frac{3\pi}{4} < -x < \frac{\pi}{4}$
- Add $\frac{\pi}{4}$: $-\frac{2\pi}{4} < \frac{\pi}{4} - x < \frac{2\pi}{4}$
- Result: $-\frac{\pi}{2} < \frac{\pi}{4} - x < \frac{\pi}{2}$
[By the property of inverse trigonometric functions, $\tan^{-1}(\tan \theta) = \theta$ if $\theta \in (-\frac{\pi}{2}, \frac{\pi}{2})$].
Step 4: Geometric Representation of the Identity
Final Simplification
Since the argument falls within the required interval, the expression simplifies directly to:
$$\tan^{-1} \left( \tan \left( \frac{\pi}{4} - x \right) \right) = \frac{\pi}{4} - x$$
Final Solution: $\frac{\pi}{4} - x$
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: $\tan^{-1} (\sqrt{\frac{1-\cos x}{1+\cos x}})$, $0 < x < \pi$
- 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: $\tan^{-1} [2\cos(2\sin^{-1}(\frac{1}{2}))]$
- 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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