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Q2:
Find the value of the following: $\tan^{-1}(\tan(\frac{7\pi}{6}))$
Solution :
Step 1: Analyzing the Domain of the Principal Value Branch
The function $f(x) = \tan^{-1}(\tan x)$ is defined such that $f(x) = x$ only if $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. [By definition of the Principal Value Branch of Inverse Tangent].
In this problem, the input value is $\theta = \frac{7\pi}{6}$. Comparing this to the domain:
- $\frac{7\pi}{6} \approx 1.166\pi$
- Upper limit of range: $\frac{\pi}{2} = 0.5\pi$
- Since $\frac{7\pi}{6} > \frac{\pi}{2}$, we cannot directly equate the expression to $\frac{7\pi}{6}$.
Step 2: Applying Trigonometric Periodicity and Reduction Formulae
The tangent function has a period of $\pi$, meaning $\tan(\theta) = \tan(\theta - n\pi)$ for any integer $n$. To bring the argument into the required range $(-\frac{\pi}{2}, \frac{\pi}{2})$, we perform the following reduction:
$\tan\left(\frac{7\pi}{6}\right) = \tan\left(\pi + \frac{\pi}{6}\right)$
[By the Quadrant Reduction Rule: $\tan(\pi + \theta) = \tan \theta$].
Therefore, $\tan\left(\frac{7\pi}{6}\right) = \tan\left(\frac{\pi}{6}\right)$.
Step 3: Evaluating the Inverse Function
Now, substitute the simplified expression back into the inverse function:
$\tan^{-1}\left(\tan\left(\frac{7\pi}{6}\right)\right) = \tan^{-1}\left(\tan\left(\frac{\pi}{6}\right)\right)$
Since $\frac{\pi}{6} \in (-\frac{\pi}{2}, \frac{\pi}{2})$, we can apply the identity $\tan^{-1}(\tan x) = x$ for $x \in (-\frac{\pi}{2}, \frac{\pi}{2})$. [By property of inverse trigonometric functions].
$\tan^{-1}\left(\tan\left(\frac{\pi}{6}\right)\right) = \frac{\pi}{6}$
Visual Representation of the Periodic Reduction
Final Solution: \frac{\pi}{6}
More Questions from Class 12 Mathematics Inverse Trigonometric Functions EXERCISE 2.MISC
- Q1: Find the value of the following: $\cos^{-1}(\cos(\frac{13\pi}{6}))$
- Q3: Prove that $2\sin^{-1} \frac{3}{5} = \tan^{-1} \frac{24}{7}$
- Q4: Prove that $\sin^{-1} \frac{8}{17} + \sin^{-1} \frac{3}{5} = \tan^{-1} \frac{77}{36}$
- Q5: Prove that $\cos^{-1} \frac{4}{5} + \cos^{-1} \frac{12}{13} = \cos^{-1} \frac{33}{65}$
- Q6: Prove that $\cos^{-1} \frac{12}{13} + \sin^{-1} \frac{3}{5} = \sin^{-1} \frac{56}{65}$
- Q7: Prove that $\tan^{-1} \frac{63}{16} = \sin^{-1} \frac{5}{13} + \cos^{-1} \frac{3}{5}$
- Q8: Prove that $\tan^{-1} \sqrt{x} = \frac{1}{2} \cos^{-1} (\frac{1-x}{1+x})$, $x \in [0, 1]$
- Q9: Prove that $\cot^{-1} (\frac{\sqrt{1+\sin x} + \sqrt{1-\sin x}}{\sqrt{1+\sin x} - \sqrt{1-\sin x}}) = \frac{x}{2}$, $x \in (0, \frac{\pi}{4})$
- Q10: Prove that $\tan^{-1}(\frac{\sqrt{1+x} - \sqrt{1-x}}{\sqrt{1+x} + \sqrt{1-x}}) = \frac{\pi}{4} - \frac{1}{2} \cos^{-1} x$, $\frac{-1}{\sqrt{2}} \leq x \leq 1$
- Q11: Solve the following equations: $2\tan^{-1}(\cos x) = \tan^{-1}(2 \csc x)$
- Q12: Solve the following equations: $\tan^{-1} \frac{1-x}{1+x} = \frac{1}{2} \tan^{-1} x, (x > 0)$
- Q13: $\sin(\tan^{-1} x)$, $|x| < 1$ is equal to
- Q14: $\sin^{-1}(1 - x) - 2\sin^{-1} x = \frac{\pi}{2}$, then $x$ is equal to
CBSE Solutions for Class 12 Mathematics Inverse Trigonometric Functions
Chapters in CBSE - Class 12 Mathematics
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